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

In this video, we will learn how to solve logarithmic equations with like bases using the laws of exponents and logarithms.

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### Video Transcript

In this video, we will learn how to solve logarithmic equations with like bases using laws of exponents and logarithms. Before we do that, let’s review how logs work. If this is a general statement for a logarithmic equation, it’s made up of a few pieces. We have logarithmic, abbreviated log. And we often say this is the log with a base 𝑏 of some value 𝑥 is equal to 𝑘. The log base 𝑏 of 𝑥 equals 𝑘 is mathematically equivalent to the statement 𝑥 is equal to 𝑏 to the 𝑘 power. The log base 𝑏 of 𝑥 is asking the question what power of 𝑏 equals 𝑥, which we see here is 𝑘. If 𝑏 to the 𝑘 power equals 𝑥, then the log base 𝑏 of 𝑥 equals 𝑘.

In order to solve problems where we’re dealing with more than one logarithm, but where these logs have the same bases, we’ll need to remember some properties of logs. We can start with the product rule. If we have some log base 𝑏 of 𝑥 times 𝑦, we can rewrite that as log base 𝑏 of 𝑥 plus log base 𝑏 of 𝑦.

The quotient rule is a bit similar. In this case, if we have log base 𝑏 of 𝑥 divided by 𝑦, we can rewrite that as log base 𝑏 of 𝑥 minus log base 𝑏 of 𝑦. From there, we have the power rule. If we have log base 𝑏 of 𝑥 to the 𝑝 power, we can rewrite that as 𝑝 times the log base 𝑏 of 𝑥. And then we have the log of one. This is the log base 𝑏 of 𝑏, and that equals one.

Remember, the log is trying to answer the question what power do we take 𝑏 to to get 𝑏. 𝑏 to the first power is equal to 𝑏. So the log base 𝑏 of 𝑏 equals one. And the final principle that will be really helpful when working with logarithms with like bases is this. If the log base 𝑏 of 𝑥 is equal to the log base 𝑏 of 𝑦, then 𝑥 equals 𝑦.

Let’s look at an example where we’re dealing with logs with the same base.

Find 𝑥 such that log base six of 𝑥 plus log base six of three is equal to three.

The first thing we notice here is that we have two log values both with a base of six. When we’re working with problems like this, the first step is usually to simplify the equation using laws of logarithms. In this equation, we’re adding two different logs with the same base, which should remind us of the product rule that tells us log base 𝑏 of 𝑥 plus log base 𝑏 of 𝑦 is equal to log base 𝑏 of 𝑥 times 𝑦. And that means we can combine these two logarithms by multiplying the values in the log. We’ll then have log base six of three times 𝑥, of three 𝑥, so that log base six of three 𝑥 equals three.

At this point, we can no longer do any simplification. So we’ll want to rewrite this log in exponential form. We remember if log base 𝑏 of 𝑥 equals 𝑘, then 𝑥 equals 𝑏 to the 𝑘 power. The base 𝑏 becomes the base of our exponential equation. The three is the exponent, which is set equal to the value we were taking the log of. Since six cubed equals 216, we have three 𝑥 equals 216. And from there, we divide both sides of the equation by three, to see that 𝑥 equals 72.

The process here is to simplify and then rewrite the equation in exponential form. This was a very simple example. But very often when we do problems like this, the simplification and the rearranging will be what we spend most of our time doing.

Let’s look at some examples where we have to do that.

Given that the log base 12 of 𝑣 plus three equals one, find the value of 𝑣.

Let’s consider this in two different ways. The first way is to rewrite this log in exponential form. If we know that log base 𝑏 of 𝑥 equals 𝑘, then we can rewrite that as 𝑏 to the 𝑘 power equals 𝑥. For us, that will be 12 to the first power equals 𝑣 plus three. We know that 12 to the first power equals 12. So we subtract three from both sides of the equation. And we see that nine equals 𝑣 or more commonly 𝑣 equals nine.

But is there another way we can think about this question? Well, if we know that log base 𝑏 of 𝑏 equals one, we could rewrite one in the form log base 12 of 12, since log base 12 of 12 does equal one. And then since we’re dealing with two logs with the same base, set equal to each other, we can say that 𝑣 plus three is equal to 12. And again, we’ll subtract three from both sides, to see that 𝑣 equals nine. Both methods are valid ways to rearrange the equation to solve for 𝑣.

Let’s consider another example.

Determine the solution set of the equation log base eight of 𝑥 minus six plus log base eight of 𝑥 plus six is equal to log base eight of 64 in all real numbers.

If we look at our equation, we see that we are adding two log values that have the same base, which means we can use the product rule log base 𝑏 of 𝑥 plus log base 𝑏 of 𝑦 equals log base 𝑏 of 𝑥 times 𝑦. However, we need to be careful here. In these logarithms, the 𝑥-value, the factor of our first log, will be 𝑥 minus six and the 𝑦-value, the factor of our second log, will be 𝑥 plus six. And this means that we’ll be dealing with the log base eight of 𝑥 minus six times 𝑥 plus six, which will be equal to log base eight of 64.

We wanna make sure we’re correctly dealing with what’s up inside the log, which means we can FOIL. Or we can recognize that this is the difference of two squares. When we have 𝑥 plus 𝑎 times 𝑥 minus 𝑎, it’s equal to 𝑥 squared minus 𝑎 squared. That would mean for us 𝑥 squared minus six squared, which equals 36. Now we have log base eight of 𝑥 squared minus 36 is equal to log base eight of 64. And once we have only one log on either side of the equal sign and they’re taken to the same base, we can say that 𝑥 squared minus 36 must be equal to 64.

To solve, we add 36 to both sides of the equation. And we see that 𝑥 squared equals 100. If we take the square root of both sides of the equation, we end up with 𝑥 is equal to plus or minus 10. But again, we’ll need to do some checks. Since our 𝑥-value goes inside a log, and we know that we can’t take the log of a negative value, if we plugged in negative 10 for 𝑥, we would be trying to find the value of a negative log, which is not possible. And so we want to say that 𝑥 in fact does not equal negative 10. And if we check positive 10, we are taking the log of positive values. And so we can say the only value that 𝑥 can be is 10. The solution set is then just 10.

In our next example, we’ll consider logs where we also have a square root.

Find the solution set of log base eight of the square root of nine 𝑥 minus 26 plus log base eight of the square root of 𝑥 plus one is equal to log base eight of 128 minus two in all real numbers.

If we copy down our equation, we see that we’re dealing with three terms that are log base eight. But one of our terms is a constant. So let’s focus on the right side of the equation, the side with the constant first. We have a few options, but it’s going to require us to think creatively.

One way of solving this is writing two as a log base eight. We do that by remembering log base 𝑏 of 𝑏 is one. And that means we could rewrite the constant two as two times log base eight of eight. That would be two times one. The other thing we remember is that if we have a constant multiplied by a log base 𝑏 of 𝑥, that’s the same thing as log base 𝑏 of 𝑥 taken to the 𝑝 power. This means the constant two can be rewritten as log base eight of eight squared, which in fact will work very well for us. So we’ll just bring down this log base eight of 128.

Now, if we have log base 𝑏 of 𝑥 minus log base 𝑏 of 𝑦, we can rewrite that as log base 𝑏 of 𝑥 divided by 𝑦. And this means we can rewrite log base eight of 128 minus log base eight of eight squared as a log base eight of 128 divided by eight squared. Eight squared is 64. 128 divided by 64 is two, which means we’ve been able to simplify the log base eight of 128 minus two down to log base eight of two.

From there, we can turn our attention to the left side of the equation. Both of our terms are log base eight. But we’re taking the log of the square root of both of these. But this will be much more helpful if we rewrite the square root as to the one-half power. Once we do that, we’re able to pull out that one-half power so that we now have one-half times the log base eight of nine 𝑥 minus 26 plus one-half times log base eight of 𝑥 plus one. Since one-half is a factor of both of these terms, we can undistribute it.

Now, inside these brackets, we have two logs with the same base that are being added together. And that means we can use the rule log base 𝑏 of 𝑥 plus log base 𝑏 of 𝑦 is equal to log base 𝑏 of 𝑥 times 𝑦. We’ll now have a log base eight of nine 𝑥 minus 26 times 𝑥 plus one. We’ll need to distribute and expand here so that we have nine 𝑥 squared plus nine 𝑥 minus 26𝑥 minus 26. Nine 𝑥 minus 26𝑥 equals negative 17𝑥.

At this point, we’re very close to something we can work with. But we don’t have exactly something we can work with yet because of this one-half. And so what we want to do is actually put this one-half back in exponent form. We want to instead call this the log base eight of nine 𝑥 squared minus 17𝑥 minus 26 to the one-half power. And hopefully, you’ll see why in just a second. Because if we do that, we now have a log base 𝑏 of 𝑥 equal to a log base 𝑏 of 𝑦, which tells us 𝑥 is equal to 𝑦, which means nine 𝑥 squared minus 17𝑥 minus 26 to the one-half power is equal to two.

We can get rid of that one-half power by squaring both sides of our equation, which gives us nine 𝑥 squared minus 17𝑥 minus 26 equals four. If we subtract four from both sides, we get the quadratic nine 𝑥 squared minus 17𝑥 minus 30 equals zero. And this is something that we can factor to solve for 𝑥.

At this point, we’re not really dealing with any log rules. We’re simply factoring to solve a quadratic. So I wanna find the terms that multiply together. Positive 10 and negative three multiply together to equal negative 30. And negative 27 plus 10 equals negative 17. Setting both of our factors equal to zero, we see that 𝑥 will either be negative ten-ninths or 𝑥 will be equal to three.

However, because of our properties of logs, we know that we can’t take the log of a negative value. If we try to plug in negative ten-ninths into nine 𝑥 minus 26, we would end up with a negative value. And that tells us that negative ten-ninths is not a valid solution for 𝑥. If we use the same method to check for three, nine times three minus 26 is positive. And if we substitute three in for 𝑥 plus one, again, we get a positive value. The solution set for 𝑥 here is just positive three.

In our final example, we’ll solve a logarithmic equation where the base is unknown.

Find the solution set of log base 𝑥 of five plus log base 𝑥 of 40 minus two times log base 𝑥 of four is equal to two plus log base 𝑥 of eight in all real numbers.

First, we’ll copy down our equation. And though there are a few strategies we could use, let’s just start by working from left to right. So we have log base 𝑏 of 𝑥 plus log base 𝑏 of 𝑦, which will be equal to log base 𝑏 of 𝑥 times 𝑦, which means we can combine our first two terms and call them log base 𝑥 of five times 40, log base 𝑥 of 200. And then we’ll just bring everything else down.

We’re close to being able to combine log base 𝑥 of 200 and two times log base 𝑥 of four. And we can use the rule that 𝑝 times log base 𝑏 of 𝑥 is equal to log base 𝑏 of 𝑥 to the 𝑝 power, which means two times log base 𝑥 of four is equal to log base 𝑥 of four squared. And then we’re subtracting two logs that have the same base. And we know when that is the case, we can divide each of their values. log base 𝑥 of 200 minus log base 𝑥 of four squared will be equal to log base 𝑥 of 200 over 16.

It seems like that’s all we can do here on the left side of the equation. What we can do now is we can move this log base 𝑥 of eight to the left side of the equation by subtracting log base 𝑥 of eight from both sides. We now have log base 𝑥 of 200 over 16 minus log base 𝑥 of eight. And we’ll use this subtraction rule again, which will be log base 𝑥 of 200 over 16 divided by eight. We know that divided by eight is the same thing as multiplied by one-eighth. And if we do that simplification, we end up with log base 𝑥 of 25 over 16 is equal to two.

And now we want to take this out of logarithm form and put it into exponent form. If we have log base 𝑏 of 𝑥 equals 𝑘, then we can rewrite that as 𝑥 equals 𝑏 to the 𝑘 power. So we have 25 over 16 is equal to 𝑥 squared. By taking the square root of both sides of the equation, we get that 𝑥 is equal to plus or minus five over four, since the square root of 25 is five and the square root of 16 is four. Since we’re only looking for values that are in the set of all reals, we want to say that 𝑥 cannot be negative five-fourths. We don’t wanna deal with a negative base here, as this would yield an answer that is imaginary. And so 𝑥 is only equal to five-fourths, which makes the solution set just five-fourths.

Before we finish, let’s review the key points. To solve logarithmic equations with like bases, first simplify the equation using law of logarithms. This might be enough to solve. If not, we can proceed to rewrite the logarithm in exponential form.