Let?s move on to our second rule of log.
So, our second rule of log stated if you have log base a, x over y
that is the same as log base a of x minus log base a of y.
Now, it's very similar to the first rule, we start off doing the same thing
makes substitutions for p so this can be p log base a of x
and this can be q log base a of y, you can then rewrite this in index form,
so let?s just say we're changing this to index form here,
so, that becomes a to the p equals to x and that becomes a to the q equals to y.
So, it's exactly the same step as you did with the first proof,
we're now starting our proof with trying to prove this part here,
so you can say that x over y, so this is x this is y and x is a to the p,
so you can rewrite this is a to the p over a to the q,
which if you?re dividing to base numbers you can rewrite this is a to the p minus q.
So, so far, we can say we have x over y equals to a to the p minus q,
you extend this a little bit so you log both sides,
so we're going to have log base a of this and log base a of this.
Looking back at our previous proof, you know that you can bring p minus q down
and this entire thing by itself will just equal to one,
so, we have log base a, x over y and if you replace your p and q, so, p is log base a of x
and q is log base a of y, we can write this as that minus log base a of y,
if we just write it out in one line here.
So, log base a, x over y as the same as log base a of x minus log base a of y.