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- How to Solve a Logarithmic Equation
- log ( 3 − x ) + log ( 4 − 3 x ) − log ( x ) = log 7 \log(3 - x) + \log (4 - 3x) - \log(x) = \log7 lo g ( 3 − x ) + lo g ( 4 − 3 x ) − lo g ( x ) = lo g 7
- 2 log 3 ( x + 4 ) − log 3 ( − x ) = 2 2 \log_3 {(x + 4)} - \log_3 (- x) = 2 2 lo g 3 ( x + 4 ) − lo g 3 ( − x ) = 2
- log 2 x = 2 + 1 2 log 2 ( x − 3 ) \log_2x = 2 + {1\over2} \log_2(x - 3) lo g 2 x = 2 + 2 1 lo g 2 ( x − 3 )
- ( log x ) 2 − log x 5 = 14 {(\log x)^2 - \log x^5 = 14} ( lo g x ) 2 − lo g x 5 = 14
- 2 ( log 3 n ) 3 − ( log 3 n ) 2 = 0 {2 (\log_3n)^3 - (\log_3n)^2 = 0} 2 ( lo g 3 n ) 3 − ( lo g 3 n ) 2 = 0
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Topic Notes
Solving logarithmic equations, rules or laws of logarithms:.
As you know, a logarithm is a mathematical operation that is the inverse of exponentiation. It is expressed by using the abbreviation "log". Before getting into solving logarithmic equations, there are several strategies and "rules" that we must first familiarize ourselves with.
First of all, in order to solve logarithmic equations, just like with polynomials, you should be comfortable graphing logarithmic functions. Check out our video on graphing logarithmic functions for an overview if needed. Also, before we get into logarithm rules, it is important that you also understand one of the simplest logarithm strategies – the change of base formula. Again, check out our video on the change of base formula if you need a refresher. Now that you have all that mastered, let's take a look at some of the most important logarithm rules:
1) Logarithm Product Rule
In general, the product rule of logarithms is defined by:
log A + log B = log ( A × B ) \log A + \log B = \log (A \times B) lo g A + lo g B = lo g ( A × B )
That is, when adding two logs of the same base , you can rewrite the expression as a single log by multiplying the terms within the logarithmic expression.
2) Logarithm Quotient Rule
In general, the quotient rule of logarithms is defined by:
log A − log B = log ( A B ) \log A - \log B = \log (\frac{A}{B}) lo g A − lo g B = lo g ( B A )
That is, when subtracting two logs of the same base , you can rewrite the expression as a single log by dividing the terms within the logarithmic expression.
3) Logarithm Power Rule
In general, the power rule of logarithms is defined by:
log ( A ) B = B × log A \log (A)^{B} = B \times \log A lo g ( A ) B = B × lo g A
That is, when there is an exponent on the term within the logarithmic expression, you can bring down that exponent and multiply it by the log.
4) Log of Exponent Rule
In general, the log of exponent rule is defined by:
log A ( A B ) = B \log_{A} (A^{B}) = B lo g A ( A B ) = B
That is, when there is an exponent on the term within the logarithmic expression, and that term is the same as the base of the logarithm, the answer is simply the exponent.
5) Exponent of Log Rule
In general, the exponent of log rule is defined by:
A log A ( B ) = B A^{\log_{A} (B)} = B A l o g A ( B ) = B
That is, raising a logarithm of a number by its base equals that number.
6) Log Identity Rule
In general, the identity rule of logarithms is defined by:
log A A = 1 \log_{A} A = 1 lo g A A = 1
That is, when taking the log of something to the base of that same thing, the logarithmic expression is simply equal to just 1.
7) Special Logs
Though not necessarily rules, there are a couple of logs that you should know by heart to make things a little easier. They are:
log 1 = 0 \log 1 = 0 lo g 1 = 0
log 0 = u n d e f i n e d \log 0 = undefined lo g 0 = u n d e f in e d
Both of these cases are always true, regardless of the base. Also, in case it comes up, the first special case is sometimes referred to as the logarithmic zero rule.
All of these rules, taken together, are extremely powerful tools we can use to solve any logarithmic problem. For a video review of these concepts, check out our videos on properties of logarithms and the quotient rule for logarithms . Now that we've covered the essentials, let's get to how to solve log problems!
How to Solve Log Problems:
As with anything in mathematics, the best way to learn how to solve log problems is to do some practice problems! We will use the rules we have just discussed to solve some examples.
Solve the logarithmic equation:
log ( 3 − x ) + log ( 4 − 3 x ) − log ( x ) = log 7 \log (3 - x) + \log (4 - 3x) - \log (x) = \log 7 lo g ( 3 − x ) + lo g ( 4 − 3 x ) − lo g ( x ) = lo g 7
Step 1: Use Known Log Rules
In any problem that involves solving logarithmic equations, the first step is to always try to simplify using the log rules. In this case, we will use the product, quotient, and exponent of log rules. We do this to try to make a polynomial/algebraic equation that is easier to solve. This is shown below:
log [ ( 3 − x ) ( 4 − 3 x ) x ] = log 7 \log [\frac{(3-x)(4-3x)}{x}] = \log 7 lo g [ x ( 3 − x ) ( 4 − 3 x ) ] = lo g 7
1 0 log [ ( 3 − x ) ( 4 − 3 x ) x ] = 1 0 log 7 10^{\log [\frac{(3-x)(4-3x)}{x}]} = 10^{\log 7} 1 0 l o g [ x ( 3 − x ) ( 4 − 3 x ) ] = 1 0 l o g 7
( 3 − x ) ( 4 − 3 x ) x = 7 \frac{(3-x)(4-3x)}{x} = 7 x ( 3 − x ) ( 4 − 3 x ) = 7
Step 2: Solve Equation
We are left with an algebraic equation which we can now solve.
3 x 2 − 13 x + 12 x = 7 \frac{3x^{2} - 13x + 12}{x} = 7 x 3 x 2 − 13 x + 12 = 7
3 x 2 − 13 x + 12 = 7 x 3x^{2} - 13x + 12 = 7x 3 x 2 − 13 x + 12 = 7 x
3 x 2 − 20 x + 12 = 0 3x^{2} - 20x + 12 = 0 3 x 2 − 20 x + 12 = 0
( 3 x − 2 ) ( x − 6 ) = 0 (3x - 2)(x - 6) = 0 ( 3 x − 2 ) ( x − 6 ) = 0
3 x − 2 = 0 o r x − 6 = 0 3x - 2 = 0 or x - 6 = 0 3 x − 2 = 0 or x − 6 = 0
x = 2 3 o r x = 6 x = \frac{2}{3} or x = 6 x = 3 2 or x = 6
Step 3: Check Solutions
Because we initially had a logarithmic equation, we need to check our answers to make sure they are valid.
log ( 3 − 2 3 ) + log ( 4 − 3 ( 2 3 ) ) − log ( 2 3 ) = log 7 \log (3 - \frac{2}{3}) + \log (4 - 3(\frac{2}{3})) - \log (\frac{2}{3}) = \log 7 lo g ( 3 − 3 2 ) + lo g ( 4 − 3 ( 3 2 )) − lo g ( 3 2 ) = lo g 7
The solution x = 2 3 x = \frac{2}{3} x = 3 2 is correct.
log ( 3 − 6 ) + log ( 4 − 3 ( 6 ) ) − log ( 6 ) = log 7 \log (3 - 6) + \log (4 - 3(6)) - \log (6) = \log 7 lo g ( 3 − 6 ) + lo g ( 4 − 3 ( 6 )) − lo g ( 6 ) = lo g 7
The solution x = 6 x = 6 x = 6 is rejected because the log of a negative number is undefined.
In this case, we will use the power and quotient log rules. We do this to try to make a polynomial/algebraic equation that is easier to solve. This is shown below:
2 log 3 ( x + 4 ) − log 3 ( − x ) = 2 2\log_{3} (x + 4) - \log_{3} (-x) = 2 2 lo g 3 ( x + 4 ) − lo g 3 ( − x ) = 2
log 3 ( x + 4 ) 2 − log 3 ( − x ) = 2 \log_{3} (x + 4)^{2} - \log_{3} (-x) = 2 lo g 3 ( x + 4 ) 2 − lo g 3 ( − x ) = 2
log 3 ( x 2 + 8 x + 16 − x ) = 2 \log_{3} (\frac{x^{2} + 8x + 16}{-x}) = 2 lo g 3 ( − x x 2 + 8 x + 16 ) = 2
Step 2: Simplify
We can convert to exponent form because one side has log and the other side does not.
3 2 = x 2 + 8 x + 16 − x 3^{2} = \frac{x^{2} + 8x + 16}{-x} 3 2 = − x x 2 + 8 x + 16
Step 3: Solve Equation
− 9 x = x 2 + 8 x + 16 -9x = x^{2} + 8x + 16 − 9 x = x 2 + 8 x + 16
x 2 + 17 x + 16 = 0 x^{2} + 17x + 16 = 0 x 2 + 17 x + 16 = 0
( x + 16 ) ( x + 1 ) = 0 (x + 16)(x + 1) = 0 ( x + 16 ) ( x + 1 ) = 0
x = − 16 o r x = − 1 x = -16 or x = -1 x = − 16 or x = − 1
Step 4: Check Solutions
2 log 3 ( ( − 16 ) + 4 ) − log 3 ( − ( − 16 ) ) = 2 2\log_{3} ((-16) + 4) - \log_{3} (-(-16)) = 2 2 lo g 3 (( − 16 ) + 4 ) − lo g 3 ( − ( − 16 )) = 2
The solution x = -16 is rejected.
2 log 3 ( ( − 1 ) + 4 ) − log 3 ( − ( − 1 ) ) = 2 2\log_{3} ((-1) + 4) - \log_{3} (-(-1)) = 2 2 lo g 3 (( − 1 ) + 4 ) − lo g 3 ( − ( − 1 )) = 2
The solution x = -1 is correct.
Step 1: Simplify
Multiply both sides of the equation by 2 to get rid of the fraction.
log 2 x = 2 + 1 2 log 2 ( x − 3 ) \log_{2} x = 2 + \frac{1}{2} \log_{2} (x - 3) lo g 2 x = 2 + 2 1 lo g 2 ( x − 3 )
2 log 2 x = 4 + log 2 ( x − 3 ) 2\log_{2} x = 4 + \log_{2} (x - 3) 2 lo g 2 x = 4 + lo g 2 ( x − 3 )
Step 2: Use Known Log Rules
In this case, we will use the power of log and quotient log rules. We can then simplify like in the previous example to make the exponential form. We do this to try to make a polynomial/algebraic equation that is easier to solve. This is shown below:
log 2 ( x ) 2 = 4 + log 2 ( x − 3 ) \log_{2} (x)^{2} = 4 + \log_{2} (x - 3) lo g 2 ( x ) 2 = 4 + lo g 2 ( x − 3 )
log 2 ( x ) 2 − log 2 ( x − 3 ) = 4 \log_{2} (x)^{2} - \log_{2} (x - 3)= 4 lo g 2 ( x ) 2 − lo g 2 ( x − 3 ) = 4
log 2 ( x 2 x − 3 ) = 4 \log_{2} (\frac{x^{2}}{x-3}) = 4 lo g 2 ( x − 3 x 2 ) = 4
2 4 = x 2 x − 3 2^{4} = \frac{x^{2}}{x-3} 2 4 = x − 3 x 2
16 = x 2 x − 3 16 = \frac{x^{2}}{x-3} 16 = x − 3 x 2
16 x − 48 = x 2 16x - 48 = x^{2} 16 x − 48 = x 2
x 2 − 16 x + 48 = 0 x^{2} - 16x + 48 = 0 x 2 − 16 x + 48 = 0
( x − 4 ) ( x − 12 ) = 0 (x - 4)(x - 12) = 0 ( x − 4 ) ( x − 12 ) = 0
x = 4 o r x = 12 x = 4 or x = 12 x = 4 or x = 12
log 2 4 = 2 + 1 2 log 2 ( 4 − 3 ) \log_{2} 4 = 2 + \frac{1}{2} \log_{2} (4 - 3) lo g 2 4 = 2 + 2 1 lo g 2 ( 4 − 3 )
The solution x = 4 checks out.
log 2 12 = 2 + 1 2 log 2 ( 12 − 3 ) \log_{2} 12 = 2 + \frac{1}{2} \log_{2} (12 - 3) lo g 2 12 = 2 + 2 1 lo g 2 ( 12 − 3 )
So does x=12. In this problem, we get to keep both our answers.
In this case, we will use the exponent of log rule. We do this to try to make a polynomial/algebraic equation that is easier to solve. Note: ( log x ) 2 (\log x)^{2} ( lo g x ) 2 is different than log x 2 \log x^{2} lo g x 2 , and thus we cannot simplify the first log \log lo g .This is shown below:
( log x ) 2 − log x 5 = 14 (\log x)^{2} - \log x^{5} = 14 ( lo g x ) 2 − lo g x 5 = 14
( log x ) 2 − 5 log x = 14 (\log x)^{2} - 5\log x = 14 ( lo g x ) 2 − 5 lo g x = 14
Step 2: Substitution
To make this equation easier to solve, we can substitute log x \log x lo g x as "a" to make a quadratic equation!
a 2 − 5 a = 14 a^{2} - 5a = 14 a 2 − 5 a = 14
a 2 − 5 a − 14 = 0 a^{2} - 5a - 14 = 0 a 2 − 5 a − 14 = 0
( a − 7 ) ( a + 2 ) = 0 (a - 7)(a + 2) = 0 ( a − 7 ) ( a + 2 ) = 0
a = 7 o r a = − 2 a = 7 or a = -2 a = 7 or a = − 2
*** Since we used substitution, we need to replace "a" back with the original term! ***
log x = 7 o r log x = − 2 \log x = 7 or \log x = -2 lo g x = 7 or lo g x = − 2
1 0 7 = x o r 1 0 − 2 = x 10^{7} = x or 10^{-2} = x 1 0 7 = x or 1 0 − 2 = x
( log 1 0 7 ) 2 − log ( 1 0 7 ) 5 = 14 (\log 10^{7})^{2} - \log (10^{7})^{5} = 14 ( lo g 1 0 7 ) 2 − lo g ( 1 0 7 ) 5 = 14
The solution x = 1 0 7 x = 10^{7} x = 1 0 7 is correct.
( log 1 0 − 2 ) 2 − log ( 1 0 − 2 ) 5 = 14 (\log 10^{-2})^{2} - \log (10^{-2})^{5} = 14 ( lo g 1 0 − 2 ) 2 − lo g ( 1 0 − 2 ) 5 = 14
The solution x = 1 0 − 2 x = 10^{-2} x = 1 0 − 2 is not correct.
And that's all there is too it! To check your work with future practice problems, be sure to use this excellent calculator here . Lastly, for a video review of everything we've just covered, check out our video on how to solve log equations .
Basic Concepts
- Evaluating logarithms using change-of-base formula
- Product rule of logarithms
- Quotient rule of logarithms
Related Concepts
- Evaluating logarithms using logarithm rules
- Graphing logarithmic functions
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Solving Logarithmic Equations
Generally, there are two types of logarithmic equations. Study each case carefully before you start looking at the worked examples below.
Types of Logarithmic Equations
- The first type looks like this.
If you have a single logarithm on each side of the equation having the same base, you can set the arguments equal to each other and then solve. The arguments here are the algebraic expressions represented by \color{blue}M and \color{red}N .
- The second type looks like this.
If you have a single logarithm on one side of the equation, you can express it as an exponential equation and solve it .
Let’s learn how to solve logarithmic equations by going over some examples.
Examples of How to Solve Logarithmic Equations
Example 1: Solve the logarithmic equation.
Since we want to transform the left side into a single logarithmic equation, we should use the Product Rule in reverse to condense it. Here is the rule, just in case you forgot.
- Apply Product Rule from Log Rules .
- Distribute: \left( {x + 2} \right)\left( 3 \right) = 3x + 6
- Drop the logs, set the arguments (stuff inside the parenthesis) equal to each other.
- Then solve the linear equation. I know you got this part down!
Just a big caution. ALWAYS check your solved values with the original logarithmic equation.
- It is OKAY for x to be 0 or negative.
- However, it is NOT ALLOWED to have a logarithm of a negative number or a logarithm of zero, 0 , when substituted or evaluated into the original logarithm equation.
CAUTION: The logarithm of a negative number, and the logarithm of zero are both not defined .
{\log _b}\left( {{\rm{negative\,\,number}}} \right) = {\rm{undefined}}
{\log _b}\left( 0 \right) = {\rm{undefined}}
Let’s check our answer to see if x=7 is a valid solution. Substitute it back into the original logarithmic equation and verify if it yields a true statement.
Yes! Since x = 7 checks, we have a solution at \color{blue}x = 7 .
Example 2: Solve the logarithmic equation.
Start by condensing the log expressions on the left into a single logarithm using the Product Rule. We want to have a single log expression on each side of the equation. Be ready though to solve for a quadratic equation since x will have a power of 2 .
- Apply Product Rule from Log Rules
- Simplify: \left( x \right)\left( {x - 2} \right) = {x^2} - 2x
- Drop the logs, set the arguments (stuff inside the parenthesis) equal to each other
- Solve the quadratic equation using the factoring method . But you need to move everything on one side while forcing the opposite side equal to 0 .
- Set each factor equal to zero, then solve for x .
x - 5 = 0 implies that x = 5
x + 2 = 0 implies that x = - 2
So the possible solutions are x = 5 and x = - 2 . Remember to always substitute the possible solutions back to the original log equation.
Let’s check our potential answers x = 5 and x = - 2 if they will be valid solutions.
After checking our values of x , we found that x = 5 is definitely a solution. However, x =-2 generates negative numbers inside the parenthesis ( log of zero and negative numbers are undefined) which makes us eliminate x =-2 as part of our solution.
Therefore, the final solution is just \color{blue}x=5 . We disregard x=-2 because it is an extraneous solution.
Example 3: Solve the logarithmic equation.
This is an interesting problem. What we have here are differences of logarithmic expressions on both sides of the equation. Simplify or condense the logs on both sides by using the Quotient Rule.
- The difference of logs is telling us to use the Quotient Rule . Convert the subtraction operation outside into a division operation inside the parenthesis. Do it to both sides of the equations.
- I think we are ready to set each argument equal to each other since we can reduce the problem to have a single log expression on each side of the equation.
- Drop the logs, and set the arguments (stuff inside the parenthesis) equal to each other. Note that this is a Rational Equation . One way to solve it is to get its Cross Product .
- It looks like this after getting its Cross Product.
- Simplify both sides by the Distributive Property. At this point, we realize that it is just a Quadratic Equation. No big deal then. Move everything to one side, which forces one side of the equation to be equal to zero.
- This is easily factorable. Now set each factor to zero and solve for x .
- So, these are our possible answers.
I will leave it to you to check our potential answers back into the original log equation. You should verify that \color{blue}x=8 is the only solution, while x =-3 is not since it generates a scenario wherein we are trying to get the logarithm of a negative number. Not good!
Example 4: Solve the logarithmic equation.
If you see “log” without an explicit or written base, it is assumed to have a base of 10 . In fact, a logarithm with base 10 is known as the common logarithm .
What we need is to condense or compress both sides of the equation into a single log expression. On the left side, we see a difference of logs which means we apply the Quotient Rule while the right side requires the Product Rule because they’re the sum of logs.
There’s just one thing that you have to pay attention to on the left side. Do you see that coefficient \Large{1 \over 2}\, ?
Well, we have to bring it up as an exponent using the Power Rule in reverse.
- Bring up that coefficient \large{1 \over 2} as an exponent (refer to the leftmost term)
- Simplify the exponent (still referring to the leftmost term)
- Then, condense the logs on both sides of the equation. Use the Quotient Rule on the left and Product Rule on the right.
- Here, I used different colors to show that since we have the same base (if not explicitly shown it is assumed to be base 10 ), it’s okay to set them equal to each other.
- Dropping the logs and just equating the arguments inside the parenthesis.
- At this point, you may solve the Rational Equation by performing Cross Product. Move all the terms on one side of the equation, then factor them out.
- Set each factor equal to zero and solve for x .
It’s time to check your potential answers. When you check x=0 back into the original logarithmic equation, you’ll end up having an expression that involves getting the logarithm of zero, which is undefined, meaning – not good! So, we should disregard or drop \color{red}x=0 as a solution.
Checking \Large{x = {3 \over 4}} , confirms that indeed \Large{\color{blue}{x = {3 \over 4}}} is the only solution .
Example 5: Solve the logarithmic equation.
This problem involves the use of the symbol \ln instead of \log to mean logarithm.
Think of \ln as a special kind of logarithm using base e where e \approx 2.71828 .
- Use Product Rule on the right side
- Write the variable first, then the constant to be ready for the FOIL method .
- Simplify the two binomials by multiplying them together.
- At this point, I simply color-coded the expression inside the parenthesis to imply that we are ready to set them equal to each other.
- Yep! This is where we say that the stuff inside the left parenthesis equals the stuff inside the right parenthesis.
- Solve the Quadratic Equation using the Square Root Method . You do it by isolating the squared variable on one side and the constant on the other. Then we apply the square root on both sides.
Don’t forget the \pm symbol.
- Simplifying further, we should get these possible answers.
Check if the potential answers found above are possible answers by substituting them back to the original logarithmic equations.
You should be convinced that the ONLY valid solution is \large{\color{blue}x = {1 \over 2}} which makes \large{\color{red}x = -{1 \over 2}} an extraneous answer.
Example 6: Solve the logarithmic equation.
There is only one logarithmic expression in this equation. We consider this as the second case wherein we have
We will transform the equation from the logarithmic form to the exponential form, then solve it.
- I color-coded the parts of the logarithmic equation to show where they go when converted into exponential form.
- The blue expression stays at its current location, but the red number becomes the exponent of the base of the logarithm which is 3 .
- Simplify the right side, {3^4} = 81 .
- Finish off by solving the two-step linear equation that arises.
You should verify that the value \color{blue}x=12 is indeed the solution to the logarithmic equation.
Example 7: Solve the logarithmic equation.
Collect all the logarithmic expressions on one side of the equation (keep it on the left) and move the constant to the right side. Use the Quotient Rule to express the difference of logs as fractions inside the parenthesis of the logarithm.
- Move all the logarithmic expressions to the left of the equation, and the constant to the right.
- Use the Quotient Rule to condense the log expressions on the left side.
- Get ready to write the logarithmic equation into its exponential form.
- The blue expression stays in its current location, but the red constant turns out to be the exponent of the base of the log.
- Simplify the right side of the equation since 5^{\color{red}1}=5 .
- This is a Rational Equation due to the presence of variables in the numerator and denominator.
I would solve this equation using the Cross Product Rule. But I have to express first the right side of the equation with the explicit denominator of 1 . That is, 5 = {\large{{5 \over 1}}}
- Perform the Cross-Multiplication and then solve the resulting linear equation.
When you check x=1 back to the original equation, you should agree that \large{\color{blue}x=1} is the solution to the log equation.
Example 8: Solve the logarithmic equation.
This problem is very similar to #7. Let’s gather all the logarithmic expressions to the left while keeping the constant on the right side. Since we have the difference of logs, we will utilize the Quotient Rule.
- Move the log expressions to the left side, and keep the constant to the right.
- Apply the Quotient Rule since they are the difference of logs.
- I used different colors here to show where they go after rewriting in exponential form.
- Notice that the expression inside the parenthesis stays in its current location, while the \color{red}5 becomes the exponent of the base.
- To solve this Rational Equation, apply the Cross Product Rule.
- Simplify the right side by the distributive property . It looks like we are dealing with a quadratic equation.
- Move everything to the left side and make the right side just zero.
Factor out the trinomial. Set each factor equal to zero then solve for x .
- When you solve for x , you should get these values of x as potential solutions.
Make sure that you check the potential answers from the original logarithmic equation.
You should agree that \color{blue}x=-32 is the only solution. That makes \color{red}x=4 an extraneous solution, so disregard it.
Example 9: Solve the logarithmic equation
I hope you’re getting the main idea now on how to approach this type of problem. Here we see three log expressions and a constant. Let’s separate the log expressions and the constant on opposite sides of the equation.
- Let’s keep the log expressions on the left side while the constant on the right side.
- Start by condensing the log expressions using the Product Rule to deal with the sum of logs.
- Then further condense the log expressions using the Quotient Rule to deal with the difference of logs.
- At this point, I used different colors to illustrate that I’m ready to express the log equation into its exponential equation form.
- Keep the expression inside the grouping symbol ( blue ) in the same location while making the constant \color{red}1 on the right side as the exponent of the base 7 .
- Solve this Rational Equation using Cross Product. Express 7 as \large{7 \over 1} .
- Cross multiply.
- Move all terms to the left side of the equation. Factor out the trinomial. Next, set each factor equal to zero and solve for x .
- These are your potential answers. Always check your values.
It’s obvious that when we plug in x=-8 back into the original equation, it results in a logarithm with a negative number. Therefore, you exclude \color{red}x=-8 as part of your solution.
Thus, the only solution is \color{blue}x=11 .
Example 10: Solve the logarithmic equation.
- Keep the log expression on the left, and move all the constants on the right side.
- I think we’re ready to transform this log equation into the exponential equation.
- The expression inside the parenthesis stays in its current location while the constant 3 becomes the exponent of the log base 3 .
- Simplify the right side since {3^3}=27 . What we have here is a simple Radical Equation .
Check this separate lesson if you need a refresher on how to solve different types of Radical Equations .
- To get rid of the radical symbol on the left side, square both sides of the equation.
- After squaring both sides, it looks like we have a linear equation. Just solve it as usual.
Check your potential answer back into the original equation.
After doing so, you should be convinced that indeed \color{blue}x=-104 is a valid solution.
You might also be interested in:
Condensing Logarithms
Expanding Logarithms
Logarithm Explained
Logarithm Rules
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How to Solve Logarithms
Last Updated: March 29, 2019 References
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Logarithms might be intimidating, but solving a logarithm is much simpler once you realize that logarithms are just another way to write out exponential equations. Once you rewrite the logarithm into a more familiar form, you should be able to solve it as you would solve any standard exponential equation.
Before You Begin: Learn to Express a Logarithmic Equation Exponentially [1] X Research source [2] X Research source
- If and only if: b y = x
- b does not equal 1
- In the same equation, y is the exponent and x is the exponential expression that the logarithm is set equal to.
- Example: 1024 = ?
- This could also be written as: 4 5
- Example: 4 5 = 1024
Method One: Solve for X
- log 3 ( x + 5) + 6 - 6 = 10 - 6
- log 3 ( x + 5) = 4
- Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 4; b = 3; x = x + 5
- Rewrite the equation so that: b y = x
- 3 4 = x + 5
- 3 * 3 * 3 * 3 = x + 5
- 81 - 5 = x + 5 - 5
- Example: x = 76
Method Two: Solve for X Using the Logarithmic Product Rule [3] X Research source [4] X Research source
- log b (m * n) = log b (m) + log b (n)
- log 4 (x + 6) + log 4 (x) = 2 - log 4 (x) + log 4 (x)
- log 4 (x + 6) + log 4 (x) = 2
- log 4 [(x + 6) * x] = 2
- log 4 (x 2 + 6x) = 2
- Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 2; b = 4 ; x = x 2 + 6x
- 4 2 = x 2 + 6x
- 4 * 4 = x 2 + 6x
- 16 = x 2 + 6x
- 16 - 16 = x 2 + 6x - 16
- 0 = x 2 + 6x - 16
- 0 = (x - 2) * (x + 8)
- x = 2; x = -8
- Example: x = 2
- Note that you cannot have a negative solution for a logarithm, so you can discard x - 8 as a solution.
Method Three: Solve for X Using the Logarithmic Quotient Rule [5] X Research source
- log b (m / n) = log b (m) - log b (n)
- log 3 (x + 6) - log 3 (x - 2) = 2 + log 3 (x - 2) - log 3 (x - 2)
- log 3 (x + 6) - log 3 (x - 2) = 2
- log 3 [(x + 6) / (x - 2)] = 2
- Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 2; b = 3; x = (x + 6) / (x - 2)
- 3 2 = (x + 6) / (x - 2)
- 3 * 3 = (x + 6) / (x - 2)
- 9 = (x + 6) / (x - 2)
- 9 * (x - 2) = [(x + 6) / (x - 2)] * (x - 2)
- 9x - 18 = x + 6
- 9x - x - 18 + 18 = x - x + 6 + 18
- 8x / 8 = 24 / 8
- Example: x = 3
Community Q&A
You Might Also Like
- ↑ http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut43_logfun.htm#logdef
- ↑ http://www.mathsisfun.com/algebra/logarithms.html
- ↑ http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut46_logeq.htm
- ↑ http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/eandl/equations/equations.html
- ↑ http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut44_logprop.htm
About This Article
To solve a logarithm, start by identifying the base, which is "b" in the equation, the exponent, which is "y," and the exponential expression, which is "x." Then, move the exponential expression to one side of the equation, and apply the exponent to the base by multiplying the base by itself the number of times indicated in the exponent. Finally, rewrite your final answer as an exponential expression. To learn how to solve for "x" in a logarithm, scroll down! Did this summary help you? Yes No
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How To Solve Logarithms in 8 Steps (With Examples)
Updated August 8, 2022
Published September 29, 2021
The Indeed Editorial Team comprises a diverse and talented team of writers, researchers and subject matter experts equipped with Indeed's data and insights to deliver useful tips to help guide your career journey.
Logarithms are mathematical calculations that use exponential functions for computation. Professionals across many career fields may use logarithms for solving various problems involving growth and decay. Solving logarithms is a beneficial skill to have when pursuing careers in research and statistics, as many roles in these sectors rely on complex mathematics.
In this article, we discuss what logarithms are, what they apply to and how to solve logs in expressions and equations.
What is a logarithm?
Logarithms are mathematical functions that transpose an exponential function to make it easier to solve. When computing algebraic equations involving exponents, logarithms serve as a valuable tool to compute unknown variables quickly while using as few mathematical components as possible. While larger logarithmic functions can require using a graphing calculator or computational software, basic logs are simple expressions that involve a few steps to compute.
A logarithm has several key elements. Looking at the example of log2_4 , the leading term "log" indicates you're working with a logarithmic function, while the subscript of 2 shows you the base. The 4 indicates the power you apply to the two. Therefore, the corresponding exponential function of the example logarithm would be 2^4, or 2 x 2 x 2 x 2 =16 . The function log2_4 equals 16. These elements are essential as you work with more complex logarithms involving unknown variables.
Uses of logarithms
Consider these important uses of logarithms across different applications:
Supports statistical research
Logarithms are an important application in many statistical, mathematical and analytical processes. Statistical research in medicine and health care, the sciences, technology, finance and business management often depends on the use of exponential analysis, which logs help simplify. Because of the range of fields that rely on mathematics, learning how to solve complex logarithmic functions can be an advantage if you're considering a career that involves data analysis and statistics.
Simplifies complex exponential problems
Exponential functions represent the rapid growth or decay of various events. In finance, exponential functions can represent costs and gains. In the sciences, environmentalists may study the growth rate of bacteria as an exponential function. For many of these types of applications, exponential functions can become challenging to solve using power notation. This makes learning how to solve logs beneficial for solving these types of functions.
Gives insight into growth and decay rates
Exponential functions are essential in measuring exponential growth and decay. The growth and decay phenomena measure how quickly something increases or decreases at a constant rate. These computations benefit from the simplification of logarithms because you can convey statistical data without the use of scientific notation and standard power form. This means you can compress large exponential values into logarithms to evaluate increases and decreases when analyzing growth and decay rates.
How to solve logs in expressions
Use the following steps to simplify logarithmic expressions:
1. Identify the base and the power
In a basic log, you can decompose the expression into its related exponential function to simplify. In the logarithm, find the base as the subscript of the log term. For instance, in the expression log7_3 , the subscript of 7 represents the base. Looking at the logarithm, identify the exponential value in the expression. This is the power you apply to the base. In the example logarithm log7_3 , the 3 is the power to which you raise the base of 7. When writing this log in exponent form, the example would be 7^3 .
2. Simplify by multiplying
Once you identify the base and power in the log, simplify the operation. Using the expression log7_3 , simplify the exponential equivalent 7^3 = 7 x 7 x 7 to get 343. This process is the most basic for solving logarithmic expressions and is essential when advancing into higher-level concepts that apply logs in algebraic equations.
3. Apply the process to larger expressions
Logarithmic expressions can get larger and include more terms, in which case you can apply the same steps as you do when solving basic functions. As an example, consider the expression (log5_3) + (log2_2) . You can decompose the function to its exponential form to get (53) + (22) . Simplifying this gives you (5 x 5 x 5) + (2 x 2) = (125 + 4) = 129 . While these examples provide numerical values that are less complex to work with, logs can include unknown values that require additional steps to this process to isolate variables.
4. Use the variable rules
When solving log expressions that have variables, apply the rules for combining like terms with exponents. For instance, the expression (loga_3) + (loga_3) + (loga_4) converts to a^3 + a^3 + a^4 . Using the rules for like terms with exponents gives the solution 2a^3 + a^ 4. This is as far as this variable expression simplifies until you find the unknown values to calculate the variable powers.
How to solve logs in equations
Use the steps below to solve logs in equations:
1. Apply the log rules to combine like terms
In logarithmic equations, you may often encounter variables that represent unknown values. The log in these problems solves for the variable to give it a numerical value. When looking at a log equation, it's important to combine like terms using the logarithm rules. These are standards that provide a framework for working through a series of operations within a log. Using the example equation log7.1_(X - 1) + log7.1_(3) = log7.1_(X + 1) and the log rules, you can combine the first two terms of the equation. This gives you log7.1_(3(X - 1)) = log7.1_(X + 1) .
2. Simplify the resulting mathematical operations
Once you combine all like terms, perform any distributive operations to multiply algebraic factors together. Using the previous example, distributing the 3 on the left side of the equal sign gives you log7.1_(3X - 3) = log7.1_(X + 1) . Once you simplify the remaining operations of a logarithm, you can isolate and solve for the variable.
3. Balance the equation to isolate the variable
Some logs carry a variable on both sides of the equal sign. In these cases, balance the equation to isolate the variable on one side of the equal sign. In the example log7.1_(3X - 3) = log7.1_(X + 1) , the variable X is on both sides of the equal sign. Set up a linear equation to show (3X - 3) = (X + 1) and solve for the variable. This gives you 2X = 4 , resulting in the solution of X = 2 . Whether you have an equation with a variable on one or both sides of the equal sign, you still isolate and solve for the unknown value.
4. Use the result to check the log
Now that you have the value for X , substitute the result in the original equation to ensure it's correct. In the original log, this looks like log7.1_(2 - 1) + log7.1_(3) = log7.1_(2 + 1) . Simplifying the equation results in log7.1_(1) + log7.1_(3) = log7.1_(3) and then log7.1_(3) = log7.1_(3) . This makes a true statement because both logarithms match on either side of the equal sign, proving the solution works.
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Properties of logarithmic functions
Comparison of exponential function and logarithmic function, practice questions, solving logarithmic functions – explanation & examples.
In this article, we will learn how to evaluate and solve logarithmic functions with unknown variables.
Logarithms and exponents are two topics in mathematics that are closely related. Therefore it is useful we take a brief review of exponents.
An exponent is a form of writing the repeated multiplication of a number by itself. An exponential function is of the form f (x) = b y , where b > 0 < x and b ≠ 1. The quantity x is the number, b is the base, and y is the exponent or power.
For example , 32 = 2 × 2 × 2 × 2 × 2 = 2 2 .
The exponential function 2 2 is read as “ two raised by the exponent of five ” or “ two raised to power five ” or “ two raised to the fifth power. ”
On the other hand, the logarithmic function is defined as the inverse function of exponentiation. Consider again the exponential function f(x) = b y , where b > 0 < x and b ≠ 1. We can represent this function in logarithmic form as:
y = log b x
Then the logarithmic function is given by;
f(x) = log b x = y, where b is the base, y is the exponent, and x is the argument.
The function f (x) = log b x is read as “log base b of x.” Logarithms are useful in mathematics because they enable us to perform calculations with very large numbers.
How to Solve Logarithmic Functions?
To solve the logarithmic functions, it is important to use exponential functions in the given expression. The natural log or ln is the inverse of e . That means one can undo the other one i.e.
ln (e x ) = x
To solve an equation with logarithm(s), it is important to know their properties.
Properties of logarithmic functions are simply the rules for simplifying logarithms when the inputs are in the form of division, multiplication, or exponents of logarithmic values.
Some of the properties are listed below.
- Product rule
The product rule of logarithm states the logarithm of the product of two numbers having a common base is equal to the sum of individual logarithms.
⟹ log a (p q) = log a p + log a q.
- Quotient rule
The quotient rule of logarithms states that the logarithm of the two numbers’ ratio with the same bases is equal to the difference of each logarithm.
⟹ log a (p/q) = log a p – log a q
The power rule of logarithm states that the logarithm of a number with a rational exponent is equal to the product of the exponent and its logarithm.
⟹ log a (p q ) = q log a p
- Change of Base rule
⟹ log a p = log x p ⋅ log a x
⟹ log q p = log x p / log x q
- Zero Exponent Rule
⟹ log p 1 = 0.
Other properties of logarithmic functions include:
- The bases of an exponential function and its equivalent logarithmic function are equal.
- The logarithms of a positive number to the base of the same number are equal to 1.
log a a = 1
- Logarithms of 1 to any base are 0.
log a 1 = 0
- Log a 0 is undefined
- Logarithms of negative numbers are undefined.
- The base of logarithms can never be negative or 1.
- A logarithmic function with base 10is called a common logarithm. Always assume a base of 10 when solving with logarithmic functions without a small subscript for the base.
Whenever you see logarithms in the equation, you always think of how to undo the logarithm to solve the equation. For that, you use an exponential function . Both of these functions are interchangeable.
The following table tells the way of writing and interchanging the exponential functions and logarithmic functions . The third column tells about how to read both the logarithmic functions.
Let’s use these properties to solve a couple of problems involving logarithmic functions.
Rewrite exponential function 7 2 = 49 to its equivalent logarithmic function.
Given 7 2 = 64.
Here, the base = 7, exponent = 2 and the argument = 49. Therefore, 7 2 = 64 in logarithmic function is;
⟹ log 7 49 = 2
Write the logarithmic equivalent of 5 3 = 125.
exponent = 3;
and argument = 125
5 3 = 125 ⟹ log 5 125 =3
Solve for x in log 3 x = 2
log 3 x = 2 3 2 = x ⟹ x = 9
If 2 log x = 4 log 3, then find the value of ‘x’.
2 log x = 4 log 3
Divide each side by 2.
log x = (4 log 3) / 2
log x = 2 log 3
log x = log 3 2
log x = log 9
Find the logarithm of 1024 to the base 2.
1024 = 2 10
log 2 1024 = 10
Find the value of x in log 2 ( x ) = 4
Rewrite the logarithmic function log 2 ( x ) = 4 to exponential form.
Solve for x in the following logarithmic function log 2 (x – 1) = 5.
Solution Rewrite the logarithm in exponential form as;
log 2 (x – 1) = 5 ⟹ x – 1 = 2 5
Now, solve for x in the algebraic equation. ⟹ x – 1 = 32 x = 33
Find the value of x in log x 900 = 2.
Write the logarithm in exponential form as;
Find the square root of both sides of the equation to get;
x = -30 and 30
But since, the base of logarithms can never be negative or 1, therefore, the correct answer is 30.
Solve for x given, log x = log 2 + log 5
Using the product rule Log b (m n) = log b m + log b n we get;
⟹ log 2 + log 5 = log (2 * 5) = Log (10).
Therefore, x = 10.
Solve log x (4x – 3) = 2
Rewrite the logarithm in exponential form to get;
x 2 = 4x – 3
Now, solve the quadratic equation. x 2 = 4x – 3 x 2 – 4x + 3 = 0 (x -1) (x – 3) = 0
Since the base of a logarithm can never be 1, then the only solution is 3.
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Solving Log Equations with Exponentials
Using the Definition Using Exponentials Calculators & Etc.
The second type of log equation requires the use of The Relationship :
—The Relationship—
...........is equivalent to............ (means the exact same thing as)
log b ( y ) = x
In animated form, the two equations are related as shown below:
Content Continues Below
MathHelp.com
Solving Logarithmic Equations
Note that the base in both the exponential form of the equation and the logarithmic form of the equation is " b ", but that the x and y switch sides when you switch between the two equations. If you can remember this — that whatever had been the argument of the log becomes the "equals" and whatever had been the "equals" becomes the exponent in the exponential, and vice versa — then you should not have too much trouble with solving log equations.
Solve log 2 ( x ) = 4
Since this equation is in the form "log(of something) equals a number", rather than "log(of something) equals log(of something else)", I can solve the equation by using The Relationship:
log 2 ( x ) = 4
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Solve log 2 (8) = x .
I can solve this by converting the logarithmic statement into its equivalent exponential form, using The Relationship:
log 2 (8) = x
But 8 = 2 3 , so I can equate powers of two:
Note that this could also have been solved by working directly from the definition of a logarithm.
What power, when put on " 2 ", would give you an 8 ? The power 3 , of course!
If you wanted to give yourself a lot of work, you could also do this one in your calculator, using the change-of-base formula:
log 2 (8) = ln(8) / ln(2)
Plug this into your calculator, and you'll get " 3 " as your answer. While this change-of-base technique is not particularly useful in this case, you can see that it does work. (Try it on your calculator, if you haven't already, so you're sure you know which keys to punch, and in which order.) You will need this technique in later problems.
I'm not saying that you'll necessarily want to solve equations using the change-of-base formula, or always by using the definition of logs, or any other particular method. But I am suggesting that you should make sure that you're comfortable with the various methods, and that you shouldn't panic if you and a friend used totally different methods for solving the same equation.
Solve log 2 ( x ) + log 2 ( x − 2) = 3
I can't do anything with this equation yet, because I don't yet have it in the "log(of something) equals a number" form. So I'll need to use log rules to combine the two terms on the left-hand side of the equation:
log 2 ( x ) + log 2 ( x − 2) = 3
log 2 [( x )( x − 2)] = 3
log 2 ( x 2 − 2 x ) = 3
Now the equation is arranged in a useful way. At this point, I can use The Relationship to convert the log form of the equation to the corresponding exponential form, and then I can solve the result:
2 3 = x 2 − 2 x
8 = x 2 − 2 x
0 = x 2 − 2 x − 8
0 = ( x − 4)( x + 2)
x = 4, −2
But if x = −2 , then " log 2 ( x ) ", from the original logarithmic equation, will have a negative number for its argument (as will the term " log 2 ( x − 2)" ). Since logs cannot have zero or negative arguments, then the solution to the original equation cannot be x = −2 .
Then my solution is:
Keep in mind that you can always check your answers to any "solving" exercise by plugging those answers back into the original equation and checking that the solution "works". In this case, I'll plug my solution value into either side of the original equation, and verify that each side evaluates to the same number:
the left-hand side:
log 2 ( x ) + log 2 ( x − 2)
= log 2 (4) + log 2 (4 − 2)3
= log 2 (4) + log 2 (2)
= log 2 (2 2 ) + log 2 (2 1 )
= 2 + 1 = 3
Solve log 2 (log 2 ( x )) = 1
This equation may look overly-complicated, but it's just another log equation. To solve this, I'll need to apply The Relationship twice. I start with the original equation and work with the "outer" log:
log 2 (log 2 ( x )) = 1
The Relationship converts the above to:
2 1 = log 2 ( x )
2 = log 2 ( x )
Now I'll apply The Relationship a second time:
Then the solution is:
Solve log 2 ( x 2 ) = (log 2 ( x )) 2
First, I'll expand the square on the right-hand side to be the explicit product of two logs:
log 2 ( x 2 ) = [log 2 ( x )] 2
log 2 ( x 2 ) = [log 2 ( x )] [log 2 ( x )]
2·log 2 ( x ) = [log 2 ( x )] [log 2 ( x )]
Then I'll move that term from the left-hand side of the equation to the right-hand side:
0 = [log 2 ( x )] [log 2 ( x )] − 2·log 2 ( x )
This equation may look bad, but take a close look. It's nothing more than a factoring exercise at this point. So I'll factor, and then I'll solve the factors by using The Relationship:
0 = [log 2 ( x )] [log 2 ( x ) − 2]
log 2 ( x ) = 0 or log 2 ( x ) − 2 = 0
2 0 = x or log 2 ( x ) = 2
1 = x or 2 2 = x
1 = x or 4 = x
You can use the Mathway widget below to practice solving logarithmic equations (or skip the widget and continue with the lesson). Try the entered exercise, or type in your own exercise. Then click the button to compare your answer to Mathway's.
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Section 6.4 : Solving Logarithm Equations
In this section we will now take a look at solving logarithmic equations, or equations with logarithms in them. We will be looking at two specific types of equations here. In particular we will look at equations in which every term is a logarithm and we also look at equations in which all but one term in the equation is a logarithm and the term without the logarithm will be a constant. Also, we will be assuming that the logarithms in each equation will have the same base. If there is more than one base in the logarithms in the equation the solution process becomes much more difficult.
Before we get into the solution process we will need to remember that we can only plug positive numbers into a logarithm. This will be important down the road and so we can’t forget that.
Now, let’s start off by looking at equations in which each term is a logarithm and all the bases on the logarithms are the same. In this case we will use the fact that,
In other words, if we’ve got two logs in the problem, one on either side of an equal sign and both with a coefficient of one, then we can just drop the logarithms.
Let’s take a look at a couple of examples.
- \(2{\log _9}\left( {\sqrt x } \right) - {\log _9}\left( {6x - 1} \right) = 0\)
- \(\log x + \log \left( {x - 1} \right) = \log \left( {3x + 12} \right)\)
- \(\ln 10 - \ln \left( {7 - x} \right) = \ln x\)
With this equation there are only two logarithms in the equation so it’s easy to get on one either side of the equal sign. We will also need to deal with the coefficient in front of the first term.
Now that we’ve got two logarithms with the same base and coefficients of 1 on either side of the equal sign we can drop the logs and solve.
Now, we do need to worry if this solution will produce any negative numbers or zeroes in the logarithms so the next step is to plug this into the original equation and see if it does.
Note that we don’t need to go all the way out with the check here. We just need to make sure that once we plug in the \(x\) we don’t have any negative numbers or zeroes in the logarithms. Since we don’t in this case we have the solution, it is \(x = \frac{1}{5}\).
Okay, in this equation we’ve got three logarithms and we can only have two. So, we saw how to do this kind of work in a set of examples in the previous section so we just need to do the same thing here. It doesn’t really matter how we do this, but since one side already has one logarithm on it we might as well combine the logs on the other side.
Now we’ve got one logarithm on either side of the equal sign, they are the same base and have coefficients of one so we can drop the logarithms and solve.
Now, before we declare these to be solutions we MUST check them in the original equation.
\(x = 6\, :\)
No logarithms of negative numbers and no logarithms of zero so this is a solution.
\(x = - 2\, :\)
We don’t need to go any farther, there is a logarithm of a negative number in the first term (the others are also negative) and that’s all we need in order to exclude this as a solution.
Be careful here. We are not excluding \(x = - 2\) because it is negative, that’s not the problem. We are excluding it because once we plug it into the original equation we end up with logarithms of negative numbers. It is possible to have negative values of \(x\) be solutions to these problems, so don’t mistake the reason for excluding this value.
Also, along those lines we didn’t take \(x = 6\) as a solution because it was positive, but because it didn’t produce any negative numbers or zero in the logarithms upon substitution. It is possible for positive numbers to not be solutions.
So, with all that out of the way, we’ve got a single solution to this equation, \(x = 6\).
We will work this equation in the same manner that we worked the previous one. We’ve got two logarithms on one side so we’ll combine those, drop the logarithms and then solve.
We’ve got two possible solutions to check here.
\(x = 2 :\)
This one is okay.
\(x = 5 :\)
This one is also okay.
In this case both possible solutions, \(x = 2\) and \(x = 5\), end up actually being solutions. There is no reason to expect to always have to throw one of the two out as a solution.
Now we need to take a look at the second kind of logarithmic equation that we’ll be solving here. This equation will have all the terms but one be a logarithm and the one term that doesn’t have a logarithm will be a constant.
In order to solve these kinds of equations we will need to remember the exponential form of the logarithm. Here it is if you don’t remember.
We will be using this conversion to exponential form in all of these equations so it’s important that you can do it. Let’s work some examples so we can see how these kinds of equations can be solved.
- \({\log _5}\left( {2x + 4} \right) = 2\)
- \(\log x = 1 - \log \left( {x - 3} \right)\)
- \({\log _2}\left( {{x^2} - 6x} \right) = 3 + {\log _2}\left( {1 - x} \right)\)
To solve these we need to get the equation into exactly the form that this one is in. We need a single log in the equation with a coefficient of one and a constant on the other side of the equal sign. Once we have the equation in this form we simply convert to exponential form.
So, let’s do that with this equation. The exponential form of this equation is,
Notice that this is an equation that we can easily solve.
Now, just as with the first set of examples we need to plug this back into the original equation and see if it will produce negative numbers or zeroes in the logarithms. If it does it can’t be a solution and if it doesn’t then it is a solution.
Only positive numbers in the logarithm and so \(x = \frac{{21}}{2}\) is in fact a solution.
In this case we’ve got two logarithms in the problem so we are going to have to combine them into a single logarithm as we did in the first set of examples. Doing this for this equation gives,
Now, that we’ve got the equation into the proper form we convert to exponential form. Recall as well that we’re dealing with the common logarithm here and so the base is 10.
Here is the exponential form of this equation.
So, we’ve got two potential solutions. Let’s check them both.
\(*x = - 2:\)
We’ve got negative numbers in the logarithms and so this can’t be a solution.
No negative numbers or zeroes in the logarithms and so this is a solution.
Therefore, we have a single solution to this equation, \(x = 5\).
Again, remember that we don’t exclude a potential solution because it’s negative or include a potential solution because it’s positive. We exclude a potential solution if it produces negative numbers or zeroes in the logarithms upon substituting it into the equation and we include a potential solution if it doesn’t.
Again, let’s get the logarithms onto one side and combined into a single logarithm.
Now, convert it to exponential form.
Now, let’s solve this equation.
Now, let’s check both of these solutions in the original equation.
\(x = - 4:\)
So, upon substituting this solution in we see that all the numbers in the logarithms are positive and so this IS a solution. Note again that it doesn’t matter that the solution is negative, it just can’t produce negative numbers or zeroes in the logarithms.
In this case, despite the fact that the potential solution is positive we get negative numbers in the logarithms and so it can’t possibly be a solution.
Therefore, we get a single solution for this equation, \(x = - 4\).
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This algebra video tutorial explains how to solve logarithmic equations ... This video include examples and practice problems with natural
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