Using the properties of logarithms: multiple steps

Video transcript

We're asked to simplify log base 5 of 25 to the x power over y. So we can use some logarithm properties. And I do agree that this does require some simplification over here, that having this right over here inside of the logarithm is not a pleasant thing to look at. So the first thing that we realize-- and this is one of our logarithm properties-- is logarithm for a given base-- so let's say that the base is x-- of a/b, that is equal to log base x of a minus log base x of b. And here we have 25 to the x over y. So we can simplify. So let me write this down. I'll do this in blue. Log base 5 of 25 to the x over y using this property means that it's the same thing as log base 5 of 25 to the x power minus log base 5 of y. Now, this looks like we can do a little bit of simplifying. It seems like the relevant logarithm property here is if I have log base x of a to the b power, that's the same thing as b times log base x of a, that this exponent over here can be moved out front, which is what we did it right over there. So this part right over here can be rewritten as x times the logarithm base 5 of 25. And then, of course, we have minus log base 5 of y. And this is useful because log base 5 of 25 is actually fairly easy to think about. This part right here is asking us, what power do I have to raise 5 to to get to 25? So we have to raise 5 to the second power to get to 25. So this simplifies to 2. So then we are left with, this is equal to-- and I'll write it in front of the x now-- 2 times x minus log base 5 of y. And we're done.