###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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# Evaluating a Logarithmic Expression in terms of Known Quantities - Problem 1

Carl Horowitz
###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Often times we will know or be given information about certain logs. So for this particular example we are told information about the log base 5 of 2 and the log base 5 of 3 and what we are asked is to express another statement in this case log base 5 of 6, in terms of these two logs.

So we are told that these are equal to j and k, so we want to get our answer in terms of j and k. Different ways you'll see this is sometimes they'll give you a numeric value so say log base 5 of 2 is equal to .4, this log base of 5 of 3 is equals to .6 and then express the same thing numerically. The process is exactly the same, instead of just having letters in here you have a number.

So what we want do when we're doing such a problem is to break down the number that we're looking for to be multiples and our factors of the numbers that we are given. So we have information about 2 and 3 and what you need to see in this case is 6 is just 2 times 3. So what we really have here is the log base 5 of 6 is equals to the log base 5 of 2 times 3.

Using our properties of logs when we are multiplying we can split that up as addition, so what this then becomes is log base 5 of 2, plus log base 5 of 3 and we know what each of these are. We know that the log base 5 of 2 is just j and the log base 5 of 3 is just k. So what we end up with is j plus k.

A common mistake is people will do this and they want to break it up as multiplication, but remember that your rules of logarithms are telling you that when you are multiplying inside the log, that turns into addition, so we actually have to end up adding our two variables.