We're asked to simplify log base 5 of x to the third.
And once again, we're just going to rewrite
this in a different way.
You could argue whether it's going to be more simple or not.
And the logarithm property that I'm
guessing that we should use for this example right here
is the property-- if I take log base x of-- let me pick
some more letters here, log base x of y to the zth power.
This is the same thing as z times log base x of y.
So this is a logarithm property.
If I'm taking the logarithm of a given base of something
to a power, I could take that power out front
and multiply that times the log of the base,
of just the y in this case.
So we apply this property over here.
And in a second, once I do this problem,
we'll talk about why this actually makes a lot of sense
and comes straight out of exponent properties.
But if we just apply that over here,
we get log base 5 of x to the third.
Well, this is the exponent right over here.
That's the same thing as z.
So that's going to be the same thing
as-- let me do this in a different color--
that 3 is the same thing-- we could put it out front-- that's
the same thing as 3 times the logarithm base 5 of x.
And we're done.
This is just another way of writing it using this property.
And so you could argue that this is a what-- maybe this
is a simplification because you took
the exponent outside of the logarithm,
and you're multiplying the logarithm by that number now.
Now with that out of the way, let's think
about why that actually makes sense.
So let's say that we know that-- and I'll just
pick some arbitrary letters here--
let's say that we know that a to the b power is equal to c.
And so if we know that-- that's written
as an exponential equation.
If we wanted to write the same truth
as a logarithmic equation, we would say logarithm base a of c
is equal to b.
To what power do I have to raise a to get c?
I raise it to the bth power.
a to the b power is equal to c.
Fair enough.
Now let's take both sides of this equation right over here,
and raise it to the dth power.
So let me make it-- so let's raise--
take both sides of this equation and raise it to the dth power.
Instead of doing it in place, I'm
just going to rewrite it over here.
So I wrote the original equation, a to the b
is equal to c, which is just rewriting this statement.
But let me take both sides of this to the dth power.
And I should be consistent.
I'll use all capital letters.
So this should be a B. Actually, let's
say I'm using all lowercase letters.
This is a lowercase c.
So let me write it this way, a to the-- so
I'm going to raise this to the dth power,
and I'm going to raise this to the dth power.
Obviously, if these two things are equal to each other,
if I raise both sides to the same power,
the equality is still going to hold.
Now, what's interesting over here
is we can now say-- what we could do
is we can use what we know about exponent properties.
Say, look, if I have a to the b power,
and then I raise that to the d power,
our exponent properties say that this is the same thing.
This is equal to a to the bd power.
This is equal to a to the bd.
Let me write it here.
This is-- let me do that in a different color.
I've already used that green.
This right over here, using what we
know about exponent properties, this
is the same thing as a to the bd power.
So we have a to the bd power is equal to c to the dth power.
And now this exponential equation,
if we would write it as a logarithmic equation,
we would say log base a of c to the dth power is equal to bd.
What power do I have to raise a to get to c to the dth power?
To get to this?
I have to raise it to the bd power.
But what do we know that b is?
We already know that b is this thing right over here.
So if we substitute this in for b,
and we can rewrite this as db, we
get logarithm base a of c to the dth power is equal to bd,
or you could also call that db, if you switch the order.
And so that's equal to d times b.
b is just log base a of c.
So there you have it.
We just derived this property.
Log base a of c to the dth, that's
the same thing as d times log base a
of c, which we applied right over here.