how to solve a logarithmic equation with different bases

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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

where ​ y ​ is an exponent, ​ b ​ is called the base and ​ x ​ is the number that results from raising the ​ b ​ to the power of ​ y ​. An equivalent expression is:

In other words, the first expression translates to, in plain English, "​ y ​ is the exponent to which ​ b ​ must be raised to get ​ x ​." For example,

because 10 3 = 1,000.

Solving problems that involve logarithms is straightforward when the base of the logarithm is either 10 (as above) or the natural logarithm ​ e ​, as these can easily be handled by most calculators. Sometimes, however, you may need to solve logarithms with different bases. This is where the change of base formula comes in handy:

This formula allows you to take advantage of the essential properties of logarithms by recasting any problem in a form that is more easily solved.

Say you are presented with the problem

Because 2 is an unwieldy base to work with, the solution is not easily imagined. To solve this type of 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

Since your calculator is equipped to solve base-10 logarithms explicitly, you can quickly find that log 50 = 1.699 and log 2 = 0.3010.

Step 3: Divide to Get the Solution

If you prefer, you can change the base to ​ e ​ instead of 10, or in fact to any number, as long as the base is the same in the numerator and the denominator.

Related Articles

Calculating logarithms, how to cancel a natural log, how to get rid of cubed power, how to put base log on graphing calculator, how to do titration calculations, how to get rid of logarithms, how to enter a subscript on the ti-83, examples of acidic buffers, how to add & multiply exponents, how to divide exponents with different bases, how to calculate log2, what is a base in chemistry, steps in learning how to do long division with bases..., acids & bases found in homes, how to convert centimeters to meters, how to calculate the area of a base, how to do powers in math, how to use log on a ti-83.

• 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

About the Author

Kevin Beck holds a bachelor's degree in physics with minors in math and chemistry from the University of Vermont. Formerly with ScienceBlogs.com and the editor of "Run Strong," he has written for Runner's World, Men's Fitness, Competitor, and a variety of other publications. More about Kevin and links to his professional work can be found at www.kemibe.com.

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

In this video, we will learn how to solve logarithmic equations involving logarithms with different bases.

Video Transcript

In this lesson, we’ll learn how to solve logarithmic equations involving logarithms with different bases. So, after this lesson, you should be able to find a solution set from an equation containing logarithms with different bases.

And we can see here what the base is of a logarithm. So, we’ve got here the base and the argument both pointed out. However, before we get on and start to solve our logarithmic equations, there’s one thing that we want to recap. And that is how we change the base of a logarithm. Well, let’s think about what we’ve got here, and this is when we’re looking at changing the base of a logarithm.

So, we’ve got log to the base 𝑎 of 𝑏. But what we have is the situation or a formula here that helps us change this to a logarithm with any base that we want. And that is that if we have log to the base 𝑎 of 𝑏, this can be equal to log to the base 𝑥 of 𝑏 over log to the base 𝑥 of 𝑎, where the base can be anything we want it to be. And we can see how that’s useful in a second cause I’ll show you an example. Well, let’s have a look at this example to see how this becomes useful.

Well, if we’ve got log to the base nine of 27, then we can rewrite this using the formula that we’ve got on the left-hand side. So, this becomes log to the base something of 27 over log to the base something of nine. And we’ve got something because it’s up to us to decide what’s going to be useful. Well, if we take a look at 27 and nine, we can see that, in fact, both of these are three to the power of something. So we’ve got three to the power of three or three squared. So therefore, we could use three as our base because this can become very useful when we have a look at one of our log laws in a second to actually simplify what we have here.

So what this is gonna give us is log to the base three of three cubed over log to the base three of three squared. And now obviously, we’ve changed the base of our logarithm here. But what I want to do is actually just show how we would simplify this. We do that by using our log rules. And the log rules we’re gonna use are that log to the base 𝑎 of 𝑚 to the power of 𝑛 equals 𝑛 log to the base 𝑎 of 𝑚. And also, log to the base 𝑎 of 𝑎 is equal to one. Well, using the first rule, we get three log to the base three of three over two log to the base three of three.

Well, then, if you use the second rule, we see that log to the base three of three is just going to be one. So, we got three multiplied by one over two multiplied by one, which would give us three over two. So, what we’ve done is we’ve shown how we changed the base of a logarithm to help us simplify cause we’ve shown that log to the base nine of 27 would be equal to three over two.

Okay, great. So now we’ve recaped some of the skills we’re gonna need to actually solve our equations. Let’s get on and look at some examples of how we solve equations using the changing of base of a logarithm.

Find the solution set of log to the base three of 𝑥 equals log to the base nine of four in the set of real numbers.

So to solve this problem, what we’re going to do is actually change the base of our logarithm. And we do that using this formula here, which is that log to the base 𝑎 of 𝑏 equals log to the base 𝑥 of 𝑏 over log to the base 𝑥 of 𝑎. And in our problem, we’re only going to change the base of the logarithm on the right-hand side of the equation. And that’s because we’re gonna keep it as log to the base three on the left-hand side because that’s what we’re gonna change the base number to with our right-hand logarithm.

Well, when we do this, we get log to the base three of 𝑥 equals log to the base three of four over log to the base three of three squared. And what I’ve done here is actually changed our nine into three squared because this, in fact, is the reason why we wanted to have the base of three. And that’s because we can actually use one of our log rules or a couple of our log rules to actually help us simplify.

Well, the first rule we’re gonna use is that log to the base 𝑎 of 𝑚 to the power of 𝑛 is equal to 𝑛 log to the base 𝑎 of 𝑚. So, we’re gonna get log to the base three of 𝑥 equals log to the base three of four over two log to the base three of three. So now what we do is apply another log rule, which tells us that log to the base 𝑎 of 𝑎 equals one.

And when we do that, we get log to the base three of 𝑥 is equal to log to the base three of four over two. That’s because we had two multiplied by one because log to the base three of three was one. So now what we can do is multiply each side by two. And when we do that, we get two log to the base three of 𝑥 equals log to the base three of four.

Well, now, if we actually apply our first rule, but in reverse, we can actually rearrange the left-hand side to be log to the base three of 𝑥 squared equals log to the base three of four. And the reason we want to do this is cause now what we can do is actually equate our arguments because we see that we’ve got log to the base three and then of 𝑥 squared and log to the base three of four. So therefore, when we do that, we’re gonna get 𝑥 squared equals four. And then when we take the square root of this, we’re gonna get 𝑥 is equal to two.

But you might think, “Well, hold on! What about the negative value, because shouldn’t we have negative two or two?” Well, in fact, we’re not interested in the negative value. And that’s because what we know is that the argument of a logarithm must be positive and not equal to one. And as we’re looking to find 𝑥, which is, in fact, the argument of the logarithm on the left-hand side, we can’t have a negative value. So therefore, we can say that the solution set of our equation is going to be two.

Okay, great. 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 equal to one. But why is that? Well, let’s take a quick look. So, as we said, if we’ve got log to the base 𝑎 of 𝑏, then we’d say that 𝑏 must be positive and not equal to one. But why is this? And you might actually apply some logic and go, “Well, hold on. If log to the base 𝑎 of 𝑏 is equal to 𝑥, then we know that 𝑎 to the power of 𝑥 is gonna be equal to 𝑏.”

Well, 𝑏 can be clearly be negative. That’s because here’s an example showing it. If we had negative three all cubed, then this would give us a result of negative 27. So why can’t 𝑏 be a negative value? Well, in fact, it’s not actually the argument that we need to look at. It’s the base, because this is what is our limiting factor and gives us our domain for the possible values for our argument. Well, in fact, we know that a base cannot be negative. And what we’re gonna do is use this example to highlight why.

So if we had log to the base negative two of 𝑥 equals a half, well, then what we’d have if we actually put it into our exponent form is negative two to the power of a half equals 𝑥. So then if we wanted to solve this for 𝑥, we’d have to take the square root of negative two. And that’s because power of a half is the same as square root. So we’d have the square root of negative two is equal to 𝑥, which we know is undefined.

So therefore, that shows why we cannot have a negative base. And therefore, following from this, if we can’t have a negative base, then we cannot have a negative argument. So that shows why it cannot be negative. So we’re also gonna show why it cannot be the value of one, and that’s because the base cannot be of value one. And that’s because let’s, for instance, think if we had log to the base one of two is equal to 𝑎, then if we rearrange this it’d be one to the power of 𝑎 equals two. Well, this can’t happen because one to the power of anything is just one. And we’re seeing the same with the example below.

Okay, great. So we now know why the argument of a logarithm must be positive and not equal to one. So let’s get on and solve some more problems.

Determine the solution set of the equation log to the base three of 𝑥 plus log to the base 243 of 𝑥 to the power of five plus three equals zero in the set of real numbers.

So to solve this problem, what we’re gonna do is use our change of base formula, which we’ve got, which is log to base 𝑎 of 𝑏 equals log to the base 𝑥 of 𝑏 over log to the base 𝑥 of 𝑎. So if we take a look at our problem, or our equation, then what we’re gonna do is we’re gonna use this change of base formula on log to the base 243 of 𝑥 to the power of five. And we’re gonna use it on this because actually what we want to do is change the base to log to the base three so we’ve got each of our logarithms with the same base.

So when we do this, what we’re gonna get is log to the base three of 𝑥 plus log to the base three of 𝑥 to the power of five over log to the base three of 243 plus three is equal to zero. And this is because we can decide what the base is when we’re changing the base. And the reason this is gonna be useful is because we know that 243 is three to the power of five. So now what we’re gonna do is use this in a couple of our log laws to help us simplify our equation.

The laws we’re gonna look at is log to the base 𝑎 of 𝑚 to the power of 𝑛 is equal to 𝑛 log to the base 𝑎 of 𝑚 and log to the base 𝑎 of 𝑎 is equal to one. So when we apply these, what we’re gonna get is log to the base three of 𝑥 is equal to five log to the base three of 𝑥, and that’s because we use our first rule for that one, and then divided by five log to the base three of three. And that’s because, as we said, 243 was the same as three to the power of five. So then we put the five in front of the log to the base three.

So we’ve got five log to the base three of three, then plus three equals zero. Well, we know that log to the base three of three is one. So therefore, we’re just gonna have five multiplied by one on the denominator. So then what we’re gonna do is divide through by five as well, which is gonna give us two log to the base three of 𝑥 plus three equals zero. And that’s cause we had log to the base three of 𝑥 and then plus log to the base three of 𝑥. So that gives us two of those. So then what we’re gonna do is subtract three from each side of the equation, which is gonna give us two log to base three of 𝑥 equals negative three.

Well, then, what we’ve got is a little trick to help us. And that is that we can turn the right-hand side into something in log to the base three. And we could do that because negative three would be the same as negative three multiplied by log to the base three of three. So therefore, we’ve got two log to the base three of 𝑥 equals negative three log to the base three of three. So then what we can do is apply the converse of our first rule. So we can have log to the base three of 𝑥 squared equals log to the base three of three to the power of negative three.

Well, because our base is the same, what we can do now is equate our arguments. So we get 𝑥 squared is equal to three to the power of negative three. So therefore, 𝑥 squared is gonna be equal to one over 27. Well, then, if we take the square root of both sides, what we’re gonna get is 𝑥 is equal to one over root 27. We’re not interested in the negative value because we’re told that we only want to find the set of real numbers. And this is because 𝑥 is, in fact, the argument in two of our logarithms, and an argument has to be positive and cannot be equal to one.

Well, then what we’re gonna do is simplify our root 27 by using one of our radical or surd rules. And that is that root 27 is same as root nine multiplied by root three, which gives us three root three. So therefore, we can say that the solution set for our equation is one over three root three.

So great, we’ve actually solved this problem with an equation. So now, let’s take a look at one that involves a quadratic.

Find the solution set of log to the base two of 𝑥 equals log to the base four of three 𝑥 plus 28 in the set of real numbers.

So in this problem, what we can do is use the change of base formula. So log to the base 𝑎 of 𝑏 equals log to the base 𝑥 of 𝑏 over log to the base 𝑥 of 𝑎. And we’re gonna use it because what we want to do is have the right-hand side with same base as the left-hand side. So when we do that, what we’re gonna get is log to the base two of 𝑥 equals log to the base two of three 𝑥 plus 28 over log to the base two of four.

And whenever we’re solving a problem like this, what we do look out for is where we could actually have something where we’ve got, for instance, log to the base two of two. So it’s the same as the form log to the base 𝑎 of 𝑎. And in fact, we can have that here because four is the same as two squared. So we can rewrite it as our denominator, being log to the base two of two squared.

So now, to simplify further, what can do is apply a couple of our log rules. First is that log to the base 𝑎 of 𝑚 to the power of 𝑛 is equal to 𝑛 log to base 𝑎 of 𝑚. And then, we’ve got log to base 𝑎 of 𝑎 equals one. So applying the first rule, we’ve got log to the base two of 𝑥 equals log to the base two of three 𝑥 plus 28 over two log to the base two of two. And then applying the second rule, we can say that the denominator becomes just two because it’s two multiplied by one, as log to the base two of two is just one.

So then, if we multiply both sides by two, we get two log to the base two of 𝑥 equals log to the base two of three 𝑥 plus 28. So now what we’re gonna do is take a look at the left-hand side and apply the converse of the first rule we looked at for our logarithms. And when we do that, we get log to the base two of 𝑥 squared equals log to the base two of three 𝑥 plus 28. So now what we can do is actually equate our arguments because we have the log to the same base on either side of the equation.

So when we do that, we now have 𝑥 squared equals three 𝑥 plus 28. So now what we can do is rearrange to actually change this into a quadratic that’s equal to zero. So we’re gonna subtract three 𝑥 and subtract 28 from each side. So, when we do that, we get the quadratic 𝑥 squared minus three 𝑥 minus 28 equals zero. So now, what we need to do is solve this for 𝑥. So, if we want to solve 𝑥 squared minus three 𝑥 minus 28 equals zero, what we’re gonna do is factor it. So, if we factor this quadratic, we’re gonna get 𝑥 minus seven multiplied by 𝑥 plus four equals zero. So therefore, we get 𝑥 is equal to seven or negative four.

So our solution set then is gonna be seven or negative four, isn’t it? Well, no because 𝑥 cannot be one of our values. The value it can’t be is negative four. And that’s because we know that if we take a look back at our equation, 𝑥 is the argument of the left-hand side. And we know that the argument must be positive and not equal to one. So therefore, the solution set of our equation is just seven.

Okay, so there we solved a problem with quadratic. What we’re gonna do now is move on to a problem where, in fact, we’re gonna have more than one result in our set.

Solve log to the base two of log to the base three of 𝑥 squared minus eight 𝑥 equals one where 𝑥 is in the set of real numbers.

So the first thing we can do with this problem is make sure that we’ve got a log to the same base on each side of the equation. And we can do that by applying one of our log rules. And that is that log to the base 𝑎 of 𝑎 equals one. So, therefore, we can say that log to the base two of log to the base three of 𝑥 squared minus eight 𝑥 equals log to base two of two because, as we said, one is just the same as log to the base two of two.

So now the reason we’ve done this is because what we’ve got is log to the same base on the left- and right-hand sides. So, what we can do is equate our arguments. So, we can say that log to the base three of 𝑥 squared minus eight 𝑥 is equal to two. So now, to solve for 𝑥, what we can do first is actually rearrange from logarithmic to exponent form. So if we have log to the base 𝑎 of 𝑏 equals 𝑥, then we say that 𝑎 to the power of 𝑥 is gonna be equal to 𝑏.

So therefore, if we identify our 𝑎, 𝑏, and 𝑥, well then we can rewrite our equation as three squared equals 𝑥 squared minus eight 𝑥, which is gonna give us nine equals 𝑥 squared minus eight 𝑥. So then, what we’re gonna do is subtract nine from each side of the equation. So, we get zero equals 𝑥 squared minus eight 𝑥 minus nine. So now what we need to do is solve this quadratic.

Well, we can solve the quadratic by factoring. And if we do, we get zero equals 𝑥 minus nine multiplied by 𝑥 plus one. So therefore, we can say that 𝑥 is gonna be equal to nine or negative one. So great, and these both are going to be in our solution set. Well, you might think, “Well, no, it can’t be because we know that the argument must be positive and not equal to one.” So therefore, we cannot have the negative value. However, that is not the case in this particular problem because if we take a look at the argument, the argument is not 𝑥. The argument is 𝑥 squared minus eight 𝑥, well, the argument of one of our logarithms.

And what we’re gonna do to show that is actually substitute in 𝑥 equals negative one into this argument. Well, if we do that, we get negative one all squared minus eight multiplied by negative one, which is gonna give us one plus eight because negative one squared is just one. And if you subtract a negative, it’s the same as adding a positive. Well, this gives us the value of nine, which is both positive and not equal to one. So therefore, it satisfies the conditions that we have for our argument. So therefore, we can say that the solution set for our equation is nine and negative one.

Okay, great. We’ve looked at a range of different problems. So now, let’s take a look at a summary of the key points. So, if we take a look at our key points, the first one is that if we want to change the base of one of our logarithms, then we can say that the log to the base 𝑎 of 𝑏 is equal to the log to the base 𝑥 of 𝑏 over the log to the base 𝑥 of 𝑎. So, we can actually have the base changed to anything that we want. And what we tend to do is choose a base that’s gonna be the most useful when we’re looking to solve an equation or simplify an expression.

Then what we also looked at is that if we have log to the base 𝑎 of 𝑏, then the 𝑏 would be our argument and the 𝑎 would be our base. And the argument must be positive and equal to one. And in fact, this is because the base must also be a positive value, not equal to one or zero. And this is important because we need to use this when considering the domain of 𝑥 if 𝑥 is part of an argument of a logarithm in an equation that we’re trying to solve.

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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.

Can someone help me?

• $\begingroup$ 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

The change of base formula is $$ \log_ab=\frac{\log b}{\log a} $$ where the base in the right hand side is whatever you prefer. I assume $e$ . The equation becomes $$ \frac{11}{\log3}\log x+\frac{7}{\log7}\log x=13+\frac{3}{\log4}\log x $$ which is a first degree equation in $\log x$ .

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

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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Unit 8: Lesson 4

• Evaluating logarithms: change of base rule

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

Check your understanding.

• 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

Challenge problems.

• (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

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