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# 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.