Question Video: Computing Logarithms by Using Laws of Logarithms Mathematics

Find the value of (log₇ 32 + log₇ 8)/(log₇ 10 − log₇ 5) without using a calculator.

02:44

Video Transcript

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.

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