Proof: log_a (B) = (log_x (B))/(log_x (A))

Proof of the logarithm property
log_a (B) = (log_x (B))/(log_x (A))

I have one last -- I was going
to say, trig property.

One last logarithm
property to show you.

So let me pick a suitably
festive color for

this last property.

So let's say that just, I don't
know, x to the n is equal to a.

Nothing fancy there.

Well, that's just another way
of saying that log base x

of a is equal to n, right?

That's the exact same -- this
is just the exact same way of

writing the exact same thing.

One's a logarithm, one's
an exponent, right?

These imply the same thing.

But what we can do is, if n is
actually equal to this

expression, we can, like I did
a couple of videos ago, you

could just substitute
this for n.

So we could write x to
this thing, log base x a.

And we could set that
as equal to what?

a.

Fascinating.

So now what I'm going to do
and, actually, this is going to

get pretty messy is, I'm going
to raise -- actually, let me

write this a little more space.

Undo.

Oh I can't keep undoing.

Anyway, so let me write
down here with more space.

Because I'm going to
do something fancy.

So, ignore this.

So, if I set x to the log base
x of a, that equals -- and

you'll see why I'm giving you
so much space right now.

Equals a.

Now, what I want to do is,
I want to raise both sides

of this equation to 1
over this exponent.

So I'm going to raise that
to 1 over log base x of a.

If I do something to one side
of the equation, I have

to do it to the other.

So that's also, that's equal to
a, to 1 over log base x to a.

I know, this is quite
daunting already.

But you'll see where I'm going.

And hopefully nothing
I've done is completely

not-intuitive, right?

This expression is just another
way of writing this expression.

And I substitut it for n.

And now I'm raising
both to this exponent.

And you'll see why
I'm doing that.

Well, if you're raising
something to an the exponent

and then you're raising that
to an exponent, you just

multiply the two, right?

So they cancel out.

Because this will
be the numerator.

And this'll be the denominator.

So that gets us to this.

x to the 1 power, right?

Because log base x of a over
log base x of a is equal to 1.

So that's the same thing
as x is equal to a to the

1 over log base x of a.

You're probably saying, Sal,
where are you going with this.

And I will sort
show you shortly.

So, we could also just replace
a with another variable, right?

I could also write x is also
equal to b to the 1 over

log base x of b, right?

Nothing strange there.

The same exact thing I did
with a, I could do with a.

The same thing I did with
a, I could do with b.

So I've written these
two expressions.

I said x is equal to
both of these things.

So let's set them
equal to each other.

So, we know that a to 1 over
log base x of a, is equal to b

to the 1 over log base x of b.

So, what can we do now?

Well let's raise both of these
-- actually, I'm running

out of so much space.

Let me clear this and go
to the next page, or

go to another page.

Clear image.

Invert.

So what did I just write?

I said that, because I
need a lot of space

for what I plan to do.

So, I said, a to the 1 over log
base x of a -- well, that

equals b to the 1 over
log base x of b.

And hopefully you're
satisfied with that.

Now, let's raise both of
these sides to the log

base x of b power.

This long base x's of b power.

Now, hopefully you'll
see why I'm doing this.

On this side they'll
cancel out, right?

Because this becomes
a numerator, that's

the denominator.

And on this side, you get a
to the -- this becomes the

numerator, right, because we
just multiply the exponents.

Log base x, that
little dot is an x.

Of b over log base x of a.

And what does that equal?

Well, that equals
just b, right?

Because this over this is 1.

This b to the 1.

That equals b.

Now let's write this entire
thing as a logarithm.

a to this thing is equal to b.

That's the exact same thing as
saying that the logarithm base

a of b is equal to this thing.

Is equal to the log base x of b
divided by the log base x of a.

This might seem confusing, it
might seem daunting, but we're

actually going to do a lot
of examples with this.

And this is probably the single
most useful identity, I

guess you could call it, if
you're using a calculator.

Why?

Because your calculator
only has two bases.

It either has log base, you
know base 10, or base e, right?

And most of them, when
you press the log button

on your calculator, it
assumes log base 10.

So if I gave you a problem
where I wanted to know what is

the log base 7 of 3, right?

Who knows?

7 to what power is 3?

And there's no easy way, on
most calculators, to do this.

Well, you can use
this identity.

That this is the same thing as
the log base 10 of 3, divided

by the log base 10 of 7.

And these are very easy to
calculate on your calculator.

You just type 3 and press log.

It'll give you this number.

And you press 7 and click
on log, it'll give

you this number.

And then you're done.

So hopefully you're satisfied
that this is true and you

have a little bit of an
intuition of how to use it.

And I'll make a bunch of videos
now, on actually how you can

use these logarithm properties.

I just wanted to get it out
of the way so that you're

satisfied that they are true.

I'll see you soon.