Lesson Video: Evaluating Logarithms Mathematics • Second Year of Secondary School

Video Transcript

In this video, we’ll learn how to evaluate logarithms of different bases using laws of logarithms.

A logarithm is a mathematical operation that determines the number of times 𝑛, a number, the base 𝑏 is multiplied by itself to get another number, 𝑚. And logarithms crop up in the most unusual places. For example, we know that musical notes vary on a logarithmic scale and that the spacing between the threads on a guitar or ukulele can be calculated using logarithms. We’re going to begin our discussion on evaluating logarithms by noting that the logarithmic function is the inverse of the exponential function. So, for example, if we consider the exponential function 𝑦 is 10 raised to the power 𝑥, if, for example, 𝑥 is equal to three, then 𝑦 is equal to 10 raised to the power of three. And that’s equal to 1000. So then, if 𝑦 is equal to 10 cubed is 1000, then the log to the base 10 of 1000 is equal to three.

So, in general, if the log to the base 𝑏 of 𝑚 is equal to 𝑛, then we say that 𝑏 raised to the 𝑛th power is equal to 𝑚, where 𝑏 is called the base of the logarithm, 𝑚 is called the argument, and 𝑛 is the exponent. And we can write this in three essentially equivalent forms. If the log to the base 𝑏 of 𝑚 is equal to 𝑛, then 𝑏 raised to the 𝑛th power is equal to 𝑚 and the 𝑛th root of 𝑚 is equal to 𝑏. And it’s important to be able to move between these forms when solving problems using logarithms.

It’s also important to know that there are two special bases of logarithms, and we see these often without the base explicitly written. If we see log with no base, then that means log to the base 10. And if we see L-N, which is ln or the natural logarithm, that means log to the base 𝑒, where 𝑒 is Euler’s number. We’ll be concentrating on the first two of our expressions. So, let’s look at some examples of how we use the equivalence between our expressions to evaluate some logarithms.

What is the value of the logarithm to the base two of eight?

The log to the base two of eight is actually the number of times we multiply two by itself to get eight. So, we’re being asked, “To what power do we raise the number two in order to get eight?” In order to find this, we’re going to use the fact that the inverse of the logarithm is the exponential. This means that if the log to the base 𝑏 of 𝑚 is equal to 𝑛, then 𝑏 raised to the 𝑛th power is equal to 𝑚. In our case, 𝑏 is equal to two, and 𝑚 is equal to eight. And we want to find the value of 𝑛.

Translating this into exponentials, this means two raised to the 𝑛th power is equal to eight. And so, we need to solve this equation for 𝑛. Now, we know that two raised to the power of one is equal to two, two squared is four, and two cubed is eight. This means then that our 𝑛 is equal to three. And so, the value of the log to the base two of eight is equal to three.

Now, let’s look at a slightly more challenging example where the base is a rational number.

What is the value of the log to the base of a half of 128?

We’re asked to find the value of a logarithm where the base is a rational number, that is, the logarithm to the base of a half of 128. To solve this, we’re going to use the fact that the logarithm and exponential functions are inverses of one another so that if log to the base 𝑏 of 𝑚 is equal to 𝑥, then 𝑏 raised to the 𝑥 power is equal to 𝑚. In our case, the base 𝑏 is equal to a half, 𝑚 is equal to 128, and we want to find the value of 𝑥. Now, we’ve already noted that our base 𝑏 is a rational number. And we know that if we raise a half to any positive power, then this will give us another fraction. So, for example, a half squared is a quarter, a half cubed is an eighth, and so on.

But we know from our correspondence between logarithms and exponentials that one over two raised to the 𝑥 power is 128. By the laws of exponents, this means one over two raised to the 𝑥 power is 128. But we know also from the laws of exponents that one over 𝑎 raised to the 𝑥 power is 𝑎 to the negative 𝑥. And this means that our exponent is going to be a negative number. Now, in order to solve our new equation, that is, two raised to the negative 𝑥 is 128, let’s look at the values of the powers of two.

We know that two raised to the power of one is two, two squared is four, and so on up to two to the power six is 64. And two raised to the seventh power is 128. And it’s this two raised to the seventh power that’s of interest to us. Because if two to the negative 𝑥 is 128 and two to the seventh power is 128, then negative 𝑥 must be equal to seven. And if negative 𝑥 is equal to seven, then our 𝑥 must be equal to negative seven. So that the value of the logarithm to the base of a half of 128 is negative seven.

Now, logarithms have some really useful properties that we can use to help us evaluate them. So, before we go on to any further examples, let’s state the rules of logarithms. The rules of logarithms apply for any base 𝑏 greater than zero, number 𝑚 greater than zero, 𝑛 greater than zero, and real number 𝑥.

Our first rule is that for any base 𝑏, the log of one is equal to zero. Our second rule says that for any base 𝑏, log to the base 𝑏 of 𝑏 is equal to one. Our third rule, which is called the power rule for logarithms, says that the log to the base 𝑏 of 𝑚 raised to the 𝑥 power is 𝑥 multiplied by log to the base 𝑏 of 𝑚. What’s happening here is that if the argument of our log has an exponent, we bring the exponent in front of our log and multiply.

Our fourth rule says that log to the base 𝑏 of 𝑚 times 𝑛 is equal to log to the base 𝑏 of 𝑚 plus log to the base 𝑏 of 𝑛. And this is the product rule for logarithms. And our fifth and final rule says, the log to the base 𝑏 of 𝑚 over 𝑛 is equal to the log to base 𝑏 of 𝑚 minus the log to the base 𝑏 of 𝑛. And that’s the quotient rule for logarithms.

We can see how these relate to the rules of exponents. We know, for example, that any positive number 𝑏 raised to the power of zero is equal to one under, and any positive number raised to the power of one is equal to itself. There is no exponent equivalent to the power rule for logarithms. However, we can see how this is derived using exponents. We know from the definition of a logarithm that log to the base 𝑏 of 𝑚 is equal to 𝑛 is equivalent to 𝑏 raised to the 𝑛th power is equal to 𝑚.

And now, if we replace our 𝑛 in the exponent for log to the base 𝑏 of 𝑚, we have 𝑏 raised to the power of the logarithm base 𝑏 of 𝑚 is equal to 𝑚. And now, we’ve raised both sides to the power 𝑛. And by the laws of exponents, 𝑎 to the power of 𝑏 raised to the power 𝑐 is equal to 𝑎 to the 𝑏 times 𝑐. We have 𝑏 raised to the power 𝑛 times log to the base 𝑏 of 𝑚 is equal to 𝑚 raised to the 𝑛th power.

But now, if we look at our expression here, which we derived from the definition of our logarithm, 𝑏 raised to the power of the logarithm to the base 𝑏 of 𝑚 to the 𝑛th power is equal to 𝑚 to the 𝑛th power. So, we have 𝑏 raised to the power of the logarithm to the base 𝑏 of 𝑚 raised to the 𝑛th power is equal to 𝑚 to the 𝑛th power is equal to 𝑏 raised to the power 𝑛 log to the base 𝑏 of 𝑚. And from the rules of exponents for 𝑎 not equal to negative one, zero, or one, if 𝑎 raised to the power of 𝑏 is equal to 𝑎 raised to the power of 𝑐, then 𝑏 must be equal to 𝑐. In our case, this means that log to the base 𝑏 of 𝑚 raised to the 𝑛th power must be 𝑛 times log to the base 𝑏 of 𝑚. That is the power rule for logarithms.

If we look now at the product rule, the product rule equivalent for exponentials is 𝑏 raised to the 𝑛th power times 𝑏 raised to the 𝑚th power is equal to 𝑏 raised to the power 𝑚 plus 𝑛. And similarly, for the quotient rule, the equivalent for exponentials is 𝑏 raised to the 𝑛th power divided by 𝑏 raised to the 𝑚th power is 𝑏 raised to the 𝑛 minus 𝑚. So now that we have our rules for logarithms, let’s use some of these in some examples.

What is the value of the logarithm to the base two of one over 128?

To find the value of the logarithm to base two of one over 128, we first note that one over 128 is equal to 128 to the power of negative one. So, this means that the logarithm to the base two of one over 128 is the logarithm to the base two of 128 to the power negative one. And now, we can apply the power rule for logarithms. This says that the log to the base 𝑏 of 𝑚 raised to the power 𝑥 is equal to 𝑥 times log to the base 𝑏 of 𝑚. And what this means is that if our argument has an exponent 𝑥, we can bring this number down and multiply our logarithm by it.

In our case, our exponent is negative one so that we have negative one times log to the base two of 128. Now, recalling our definition of a logarithm, if the logarithm to the base 𝑏 of 𝑚 is equal to 𝑛, then 𝑏 raised to the power 𝑛 is equal to 𝑚. And what this means is that 𝑛 is the number of times we multiply the base 𝑏 by itself to give 𝑚. So now, we’re looking for the negative of the number of times we multiply the base two by itself to get 128. So, we need to find a value of 𝑛 for which two raised to the 𝑛th power is 128.

And if we look at our powers of two, we can see that two to the seventh power is 128. So that our 𝑛 is equal to seven. This means that log to the base two of 128 is equal to seven. Therefore, negative log to the base two of 128 is negative seven. And we have our solution log to the base two of one over 128 is negative seven.

It’s worth noting, perhaps, that we could have done this in a slightly different way, using the fact that two to the power of seven is 128. Writing log to the base two of one over 128 is equal to negative log to the base two of 128, we can write this as negative log to the base two of two raised to the power of seven. And using our power rule for logarithms once more, we get negative seven log to the base two of two. We know from our laws of logarithms that log to the base 𝑏 of 𝑏 is equal to one. And we have here log to the base two of two, so that’s equal to one. So, again, we reach our answer of a negative seven.

Now, let’s look at an example where we use a combination of the power and the product rules to calculate logarithms.

Calculate two log four plus seven log 13, giving your answer to the nearest thousandth.

We’re asked to calculate two log four plus seven log 13. We can begin by using the power rule for logarithms. This says that 𝑥 times the logarithm to the base 𝑏 of 𝑚 is equal to the logarithm to the base 𝑏 of 𝑚 raised to the power 𝑥. In our case, we can use this twice. In our first expression two log four, our 𝑥 is two, and 𝑚 is four. So by raising 𝑚 to the power 𝑥, we can rewrite this as the log of four squared. And then for our second expression seven log 13, where our 𝑥 is seven and 𝑚 is 13, we havethe log of 13 raised to the power seven.

It’s worth reminding ourselves here that if the logarithm is written without a base, which ours is, this means that the base is 10. And so, we have two log four plus seven log 13 is log four squared plus log 13 to the power seven, where all of our logarithms are to the base 10. Now, to simplify our expression on the right-hand side, we’re going to use the product rule for logarithms. This says that the log to the base 𝑏 of 𝑚 times 𝑛 is equal to log to the base 𝑏 of 𝑚 plus log to the base 𝑏 of 𝑛.

And with 𝑚 is four squared and 𝑛 equals 13 raised of the power seven, we have two log four plus seven log 13 is equal to the log of four squared times 13 raised to the power seven. And now, we simply need to use the log button on our calculator to evaluate this. This gives us 9.00172 and so on, which to the nearest one thousandth is 9.002. And so to the nearest one thousandth, we have two log four plus seven log 13 is 9.002.

In our final example, we’ll use a combination of the rules of logarithms to evaluate a logarithmic expression.

Find the value of the logarithm to the base two of log of 𝑥 raised to the power 32 minus log to the base two of log of 𝑥 raised to the power four.

We’re asked to find the value of what looks like a complicated log expression. We have a logarithm to the base two of another logarithm to the base 10. Recalling that a logarithm written without the base is log to the base 10, and this is the logarithm of 𝑥 raised to power 32. And our second expression is similar. However, notice that what we have is logarithm to the base two minus another logarithm to the base two. And this means we can use the quotient rule for logarithms. This says that log to the base 𝑏 of 𝑚 minus log to the base 𝑏 of 𝑛 is equal to log to the base 𝑏 of 𝑚 divided by 𝑛.

So that in our case where our base 𝑏 is equal to two, we have log to the base two of the log of 𝑥 raised to the power 32 divided by log of 𝑥 raised to the power four. And for each of the expressions in our quotient, we can use the power rule for logarithms. This says that log to the base 𝑏 of 𝑚 raised to the power 𝑥 is equal to 𝑥 times the log to the base 𝑏 of 𝑚. That is, if our argument has an exponent, we simply multiply by the exponent. In our case then, our exponents are 32 and four. And if we bring these down as our argument, we have 32 log 𝑥 divided by four log 𝑥. log 𝑥 divided by log 𝑥 is equal to one. So, we simply have log to the base two of 32 over four, which is, of course, log to the base two of eight.

And now, we can use the definition of our logarithm to find the value of this. That is, if the log to the base 𝑏 of 𝑚 is equal to 𝑛, then 𝑏 raised to the power 𝑛 is equal to 𝑚. So that 𝑛 is the number of times 𝑏 the base is multiplied by itself to give 𝑚. In our case, the base 𝑏 is two, and 𝑚, the argument, is eight. So, we need to know how many times does two multiply by itself to give eight or to what power 𝑛 we raise two to get eight. Now, we know that two times two times two, which is two cubed, is equal to eight. So that our 𝑛 is equal to three. And we have log to the base two of the log of the 𝑥 raised of the power 32 minus log to the base two of log of 𝑥 to the power four is equal to three.

Notice that we could have also used the power rule in the final steps to get our answer. Since log to the base two of eight is equal to log to the base two of two to the power of three. And by the power rule that’s equal to three times log to the base two of two. And since log to the base 𝑏 of 𝑏 is equal to one for any base 𝑏, our log to the base two of two is equal to one. And we arrive again at our answer three.

Let’s finish by reminding ourselves of some key points for evaluating logarithms. Remember that a logarithm to the base 𝑏 of 𝑚 which is equal to 𝑛 is the number of times 𝑛 that the base must be multiplied by itself to obtain the argument 𝑚. The logarithm function is the inverse of the exponential function. So that if the logarithm to the base 𝑏 of 𝑚 is equal to 𝑛, then 𝑏 raised to the power 𝑛 is equal to 𝑚. The two cases where the bases are not stated, that’s log, which means log to the base 10, and L-N or ln is log to the base 𝑒, the natural logarithm. And we can use the rules of logarithms to simplify, calculate, and evaluate logarithms and logarithmic expressions.

Where for the base 𝑏 greater than zero, 𝑚 greater than zero, 𝑛 greater than zero, and 𝑥 real number, our rules are log to the base 𝑏 of one is equal to zero, log to the base 𝑏 of 𝑏 is equal to one, log to the base 𝑏 of 𝑚 raised to the power 𝑥 is equal to 𝑥 times log to the base 𝑏 of 𝑚. And that’s the power rule. log to the base 𝑏 of 𝑚 times 𝑛 is equal to log to the base 𝑏 of 𝑚 plus log to the base 𝑏 of 𝑛. And that’s the product rule. And finally, log to the base 𝑏 of 𝑚 over 𝑛 is equal to the log 𝑏 of 𝑚 minus log to the base 𝑏 of 𝑛. And that’s the quotient rule.