# Lesson Video: Laws of logarithms Mathematics

In this video, we will learn how to use the laws of logarithms to simplify logarithmic expressions and prove logarithmic statements.

15:15

### Video Transcript

In this video, we’ll learn how to use the laws of logarithms to simplify logarithmic expressions. We’ll begin by recalling the definition of a logarithm and how these link with exponentiation. We’ll then use this link to develop a few of the laws of logarithms that are gonna help us to simplify logarithmic expressions.

Let’s begin by looking at a single power of two. We know that two to the fifth power is equal to 32. Another way to think about this is what power of two gives us 32. We can write this in logarithmic form as log base two of 32 is equal to what. Well, we already saw that it’s equal to five. These two equations are equivalent ways of expressing the same relationship.

In general, let’s take a positive real number, 𝑏, which is not equal to one and two other positive numbers, 𝑥 and 𝑦. We say that log base 𝑏 of 𝑦 is equal to 𝑥 is equivalent to saying 𝑏 to the power of 𝑥 is equal to 𝑦. We say that 𝑏 is the base, 𝑥 is the exponent, and 𝑦 is the argument.

Now, we’re interested in the laws of logarithms and how these can help us to simplify algorithmic expressions. And since a logarithm is an equivalent way of expressing the relationship between a number and its exponent, we’re able to rewrite our exponent laws. The first exponent law that we recall is that 𝑏 to the power of 𝑥 one times 𝑏 to the power of 𝑥 two is equal to 𝑏 to the power of 𝑥 one plus 𝑥 two. In other words, for real constants 𝑏, as long as the base is the same, we simply add the exponents.

The equivalent law of logarithms says that log base 𝑏 of 𝑥 one times 𝑥 two is equal to log base 𝑏 of 𝑥 one plus log base 𝑏 of 𝑥 two. Similarly, we compare the law of exponents for dividing to the law of logarithms. And we find log base 𝑏 of 𝑥 one over 𝑥 two is log base 𝑏 of 𝑥 one minus log base 𝑏 of 𝑥 two.

Now, it’s outside the scope of this video to prove all of these. But continuing in this manner, and we find a law for dealing with powers. That is, log base 𝑏 of 𝑥 to the power of 𝑝 is the same as 𝑝 times log base 𝑏 of 𝑥. And then we have the change of base formula. That is, log base 𝑏 of 𝑥 one over log base 𝑏 of 𝑥 two is the same as log base 𝑥 two of 𝑥 one. It’s also worth noting the following two laws. Log base 𝑏 of 𝑏 is one and log base 𝑏 of one is zero. What we’re really interested in is looking at how we can apply these laws to simplify logarithms.

Calculate log base two of 192 minus log base two of three.

Let’s begin by recalling a suitable law of logarithms. This says that for a fixed base 𝑏 is greater than zero which is not equal to one and positive numbers 𝑥 one and 𝑥 two, log base 𝑏 of 𝑥 one divided by 𝑥 two is the same as log base 𝑏 of 𝑥 one minus log base 𝑏 of 𝑥 two. The converse is true. So we say that when subtracting logarithms with the same base, we simply divide their arguments.

In our case, we can let 𝑏 be equal to two, 𝑥 sub one is 192, and 𝑥 sub two is equal to three. And we can, therefore, say that log base two of 192 minus log base two of three is the same as log base two of 192 divided by three. 192 divided by three is 64. So we can write this as log base two of 64. Well, we’ve simplified, but we still need to calculate this logarithm. In other words, we need to work out what log base two of 64 actually is. And we, therefore, need to recall the definition of a logarithm.

In general, we say that log base 𝑏 of 𝑦 equals 𝑥 is equivalent to saying 𝑏 to the power of 𝑥 equals 𝑦. And so, essentially, we’re asking what exponent of two gives 64. Well, we know that two to the power of six is 64. And so log base two of 64 must be equal to six. And so the answer here is six.

Let’s consider another example.

Find the value of log base two of 10 plus log base two of 16 minus log base two of five without using a calculator.

Firstly, we’re going to recall the order of operations. That tells us that when there is an addition and a subtraction in the same sum, we simply move from left to right. So we’ll begin by evaluating log base two of 10 plus log base two of 16. And then we’ll subtract log base two of five. And so we’re going to recall some laws of logarithms.

The first is sometimes called the product law. And this says for a fixed base 𝑏 which is greater than zero and not equal to one and positive numbers 𝑥 one and 𝑥 two, log base 𝑏 of 𝑥 one times 𝑥 two is log base 𝑏 of 𝑥 one plus log base 𝑏 of 𝑥 two. Of course, the converse is true. So we can say that to add logarithms whose base is the same, we simply multiply the argument.

Similarly, with quotients, log base 𝑏 of 𝑥 one divided by 𝑥 two is log base 𝑏 of 𝑥 one minus log base 𝑏 of 𝑥 two. And so we can say that log base two of 10 plus log base two of 16 is equal to log base two of 10 times 16. Usually, we’d look to evaluate that, but we’re not going to just yet. Instead, we’re going to move straight on to subtracting log base two of five. And we know this means we need to divide the arguments. So we get log base two of 10 times 16 divided by five.

And we now see that since we didn’t simplify, we can divide both the numerator and denominator of our fraction by five, giving us log base two of two times 16 over one, which is log base two of 32. Remember, we’re trying to find the value of this logarithm. So we recall the definition of a logarithm. We say that log base 𝑏 of 𝑦 equals 𝑥 is equivalent to saying 𝑏 to the power of 𝑥 equals 𝑦. Well, here, our base is two. So we’re essentially asking what exponent of two will give us 32. Well, we know that two to the fifth power is 32, and so log base two of 32 must be five. The value of log base two of 10 plus log base two of 16 minus log base two of five is five.

We’ll now consider an example that requires the use of the change of base formula.

Find the value of log base seven of 32 plus log base seven of eight divided by log base seven of 10 minus log base seven of five without using a calculator.

Let’s recall some of the laws of logarithms. We know that when adding logarithms whose base is equal, we simply multiply the argument. So log base 𝑏 of 𝑥 one plus log base 𝑏 of 𝑥 two is log base 𝑏 of 𝑥 one times 𝑥 two. We have a similar rule for subtracting, but this time we divide the arguments. And so let’s use these rules to evaluate the numerator and denominator of our fraction. Log base seven of 32 plus log base seven of eight is the same as log base seven of 32 times eight, but 32 times eight is 256. So our numerator becomes log base seven of 256. Then our denominator is log base seven of 10 divided by five, which is log base seven of two. And so we’ve simplified a little bit, and our fraction becomes log base seven of 256 divided by log base seven of two.

Now, we need to be really careful here. A common mistake is to think that because when we divide the arguments we subtract the two logarithms, we can simply subtract these values. Remember, that’s not actually what our log laws say. Instead, we’re going to apply the change of base formula, so called because it literally allows us to change the base that we’re working with.

For this to work, we need to have a fraction made up of two logarithms whose base is the same. So log base 𝑏 of 𝑥 one divided by log base 𝑏 of 𝑥 two is then log base 𝑥 two of 𝑥 one. Comparing this general form with our fraction, we find the base 𝑏 is equal to seven. 𝑥 one is the argument of the logarithm on the top of our fraction, so it’s 256. And 𝑥 sub two is the argument of the logarithm on our denominator, so it’s two. This means that we can now write log base seven of 256 over log base seven of two as log base two of 256.

We’re still not finished though. We have fully simplified it, but we need to evaluate this. And so let’s recall the definition of a logarithm. If we say log base 𝑏 of 𝑦 equals 𝑥, we can equivalently say that 𝑏 to the power of 𝑥 must be equal to 𝑦. And so, here, since our base is two, we’re asking, what exponent of two gives us a value of 256? Well, two to the eighth power is 256. And this then means that log base two of 256 must be eight. Log base seven of 32 plus log base seven of eight all divided by log base seven of 10 minus log base seven of five is eight.

In our next example, we’ll look at how to approach a problem where the base isn’t explicitly written.

Which of the following is equal to five log three over log four plus log six?

This expression might look a little bit strange as our logs seem to have no base. If a log has no base, we generally assume that the base is equal to 10. And so we rewrite our fraction as five log base 10 of three over log base 10 of four plus log base 10 of six.

So we’re next going to recall some rules for logarithms. Firstly, we know that log base 𝑏 of 𝑥 one plus log base 𝑏 of 𝑥 two is log base 𝑏 of 𝑥 one times 𝑥 two. As long as our bases are the same, we simply multiply the arguments. And so the denominator of our fraction is going to become log base 10 of four times six, which is log base 10 of 24.

And what about our numerator? Well, log base 𝑏 of 𝑥 to the power of 𝑝 for real constants 𝑝 is the same as 𝑝 times log base 𝑏 of 𝑥. The converse is true. So we can write our denominator as log base 10 of three to the fifth power. But three to the fifth power is 243. And so we have log base 10 of 243 over log base 10 of 24.

Note that we have a fraction with two logarithms whose bases are equal. And so we can use the change of base formula. This says that log base 𝑏 of 𝑥 one divided by log base 𝑏 of 𝑥 two can be written as log base 𝑥 two of 𝑥 one. So, essentially, if the bases are the same, we make the argument of our denominator the new base. And the argument of our numerator becomes the new argument. In this case then, the base of our logarithm is 24 and its new argument is 243. So we can write our fraction as log base 24 of 243. And our correct answer is therefore (C).

Note that this actually means it didn’t matter what base we assumed. Because we ended up with two logarithms with the same base, we simply apply the change of base formula. We could’ve chosen base two or base three. But remember, the general convention is to assume that it’s base 10.

We’re going to consider one final example of how we might apply the change of base formula.

Simplify log base three of 16 times log base two of 243.

We need to be really, really careful here. The first common mistake would be to confuse the laws of logarithms and think we can add 16 and 243. That, of course, is the opposite of the rule for products and would not be applicable either way, since the bases are actually different here. And so, instead, we’re going to recall the change of base formula. This says that log base 𝑏 of 𝑥 one divided by log base 𝑏 of 𝑥 two can be written as log base 𝑥 two of 𝑥 one. We need a fraction with logarithms in the numerator and denominator. The base of these logarithms needs to be equal. And if this is the case, we can rewrite this as a logarithm with a new base of the argument of our denominator.

Now, in fact, we’re finding the product of two logarithms, so we’re going to rewrite our formula. We’re going to multiply both sides of the equation by log base 𝑏 of 𝑥 two. And then we’re going to compare this to our expression. Note that in the product, the base of the first log is equal to the argument of the second. Now, this doesn’t quite work. We don’t have a common value for 𝑥 two in our expression, but we can rewrite our first logarithm. We know 16 is two to the fourth power. So we rewrite log base three of 16 as log base three of two to the fourth power. And then we use our exponent rule and we write this as four times log base three of two. So our expression now becomes four log base three of two times log base two of 243.

Since multiplication is commutative, it can be performed in any order, we can add parentheses. And we’ll evaluate log base three of two times log base two of 243 first. We’re going to let 𝑥 two be equal to two. Then 𝑥 one is 243 and 𝑏 is equal to three. This means we can write log base three of two times log base two of 243 as log base three of 243. And our expression is now four times log base three of 243. But actually, we can evaluate log base three of 243. By recalling the definition of the logarithm, we ask ourselves, well, what exponent of three gives 243?

We know three to the fifth power is 243. So log base three of 243 is actually equal to five. And so our expression simplifies really nicely to four times five, which is equal to 20. Log base three of 16 times log base two of 243 by using the change of base formula is 20.

In this video, we learned that for a fixed base 𝑏 which is greater than zero and not equal to one and positive numbers 𝑥 one, 𝑥 two, and 𝑥, log base 𝑏 of 𝑥 one plus log base 𝑏 of 𝑥 two is log base 𝑏 of 𝑥 one times 𝑥 two. Similarly, log base 𝑏 of 𝑥 one minus log base 𝑏 of 𝑥 two is log base 𝑏 of 𝑥 one divided by 𝑥 two. And for real constants 𝑝, log base 𝑏 of 𝑥 to the power of 𝑝 is the same as 𝑝 times log base 𝑏 of 𝑥. We saw that the change of base formula tells us that log base 𝑏 of 𝑥 one divided by log base 𝑏 of 𝑥 two is log base 𝑥 two of 𝑥 one. We note that log base 𝑏 of 𝑏 equals one and log base 𝑏 of one equals zero and that if a logarithmic expression has no base, we assume its base to be equal to 10.