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## How to Solve Logarithms With Different Bases

## How to Evaluate Logarithms With Square Root Bases

A logarithmic expression in mathematics takes the form

Say you are presented with the problem

## Step 1: Change the Base to 10

Using the change of base formula, you have

This can be written as log 50/log 2, since by convention an omitted base implies a base of 10.

## Step 2: Solve for the Numerator and Denominator

## Step 3: Divide to Get the Solution

## Related Articles

- Mesa Community College: Using the Change Of Base Property to Evaluate Logarithms
- Michigan State University: Change-of-Base Formula
- Khan Academy: Logarithm Change of Base Rule Intro

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## Lesson Video: Logarithmic Equations with Different Bases Mathematics • 10th Grade

## Video Transcript

## Stack Exchange Network

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## How do I solve this logarithm equation with different bases?

How do is solve this logarithm equation?

$$11 \cdot \log_3x+7 \cdot \log_7x = 13+3 \cdot \log_4x$$

I know that I have to use the change of base formula, but I still can't figure out the equation.

- $\begingroup$ Is that eleven times $\log_3x$, or 11 times $x$ times $\log_3 x$? Very different things... $\endgroup$ – Arturo Magidin Feb 15, 2019 at 17:42
- $\begingroup$ But the image in the original question was $11x\log_3x + \cdots$. Here's the image: i.stack.imgur.com/q0rTG.png $\endgroup$ – stressed out Feb 15, 2019 at 17:46
- $\begingroup$ Which is it? The image you posted has an $x$, not a product. Such an equation would be much more difficult to solve than the one you have currently, with $11\log_3x$ instead of $11x\log_3 x$. Please make sure you are asking about the correct equation you are expected to solve. $\endgroup$ – Arturo Magidin Feb 15, 2019 at 18:07
- $\begingroup$ With the x I meant a product. Sorry for the confusion. $\endgroup$ – sarina eevers Feb 15, 2019 at 18:13

## 2 Answers 2

You may want to use the identity: $\log_{a}(x)=\frac{\log(x)}{\log(a)}$

Thus $11 \cdot \log_3x+7 \cdot \log_7x = 13+3 \cdot\log_4x$ becomes:

$\frac{11\log(x)}{\log(3)}+\frac{7\log(x)}{\log(7)}=13+3\frac{\log(x)}{\log(4)}$ .

Then set $u=\log(x)$ and solve for $u$ , and then find $x$ .

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## Solving Logarithmic Equations

## Types of Logarithmic Equations

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.

- 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}}

Yes! Since x = 7 checks, we have a solution at \color{blue}x = 7 .

Example 2: Solve the logarithmic equation.

- 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 + 2 = 0 implies that x = - 2

Let’s check our potential answers x = 5 and x = - 2 if they will be valid solutions.

Example 3: Solve the logarithmic equation.

- 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.

Example 4: Solve the logarithmic equation.

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 .

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.

Example 6: Solve the logarithmic equation.

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.

Example 7: Solve the logarithmic equation.

- 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.

Example 8: Solve the logarithmic equation.

- 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 .

Make sure that you check the potential answers from the original logarithmic equation.

Example 9: Solve the logarithmic 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.

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 .

- 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.

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## Unit 8: Lesson 4

## Logarithm change of base rule intro

- Using the logarithm change of base rule
- Proof of the logarithm change of base rule
- Logarithm properties review

## The change of base rule

- When using this property, you can choose to change the logarithm to any base x \greenE x x start color #0d923f, x, end color #0d923f .
- As always, the arguments of the logarithms must be positive and the bases of the logarithms must be positive and not equal to 1 1 1 1 in order for this property to hold!

## Example: Evaluating log 2 ( 50 ) \log_2(50) lo g 2 ( 5 0 ) log, start base, 2, end base, left parenthesis, 50, right parenthesis

- Your answer should be
- an integer, like 6 6 6 6
- a simplified proper fraction, like 3 / 5 3/5 3 / 5 3, slash, 5
- a simplified improper fraction, like 7 / 4 7/4 7 / 4 7, slash, 4
- a mixed number, like 1 3 / 4 1\ 3/4 1 3 / 4 1, space, 3, slash, 4
- an exact decimal, like 0.75 0.75 0 . 7 5 0, point, 75
- a multiple of pi, like 12 pi 12\ \text{pi} 1 2 pi 12, space, start text, p, i, end text or 2 / 3 pi 2/3\ \text{pi} 2 / 3 pi 2, slash, 3, space, start text, p, i, end text

## Justifying the change of base rule

- (Choice A) log ( a ) \log(a) lo g ( a ) log, left parenthesis, a, right parenthesis A log ( a ) \log(a) lo g ( a ) log, left parenthesis, a, right parenthesis
- (Choice B) log ( 6 ) \log(6) lo g ( 6 ) log, left parenthesis, 6, right parenthesis B log ( 6 ) \log(6) lo g ( 6 ) log, left parenthesis, 6, right parenthesis
- (Choice C) log ( 6 a ) \log(6a) lo g ( 6 a ) log, left parenthesis, 6, a, right parenthesis C log ( 6 a ) \log(6a) lo g ( 6 a ) log, left parenthesis, 6, a, right parenthesis
- (Choice D) log 6 ( 6 a ) \log_{6}(6a) lo g 6 ( 6 a ) log, start base, 6, end base, left parenthesis, 6, a, right parenthesis D log 6 ( 6 a ) \log_{6}(6a) lo g 6 ( 6 a ) log, start base, 6, end base, left parenthesis, 6, a, right parenthesis

## IMAGES

## VIDEO

## COMMENTS

This algebra 2 and precalculus video tutorial focuses on solving logarithmic equations with different bases. To do this, you need to

How do we solve a log equation with different bases? Here we will see how we can use the change of base formula for logarithm to solve

Learn how to Solve Logarithmic Equations with different bases using the Change of Base Formula. Also check your answer for extraneous

How to Solve Logarithms With Different Bases · Step 1: Change the Base to 10 · Step 2: Solve for the Numerator and Denominator · Step 3: Divide to

In order to solve the logarithmic equation, we will make use of the change in base formula: l o g l o g l o g 𝑥 = 𝑥 𝑎 . If we use the base 𝑏 = 𝑥

So that's our first equation solved. But here, we did mention something. We mentioned that the argument of a logarithm must be positive and not

The change of base formula is logab=logbloga. where the base in the right hand side is whatever you prefer. I assume e. The equation becomes

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.

Solve the equation log (3 x + 1) = 1. Use like bases to solve the exponential equation. 5^(-8v - 7) = 5^(-v). ... Solve the following equation: \log (3x) + \log(5)

By the same logic, we can prove the change of base rule. Just change 2 2 2