SOLUTION: How do you get from {{{log(a, ( (x-y)/(sqrt(x^2-y^2)) ))}}} to {{{1/2}}}{{{(log(a, (x-y))-log(a,(x+y))^"") }}}?

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How do you get from {{{log(a, ( (x-y)/(sqrt(x^2-y^2)) ))}}} to {{{1/2}}}{{{(log(a, (x-y))-log(a,(x+y))^"") }}}?      Log On


   

Question 301241: How do you get from log%28a%2C+%28++%28x-y%29%2F%28sqrt%28x%5E2-y%5E2%29%29++%29%29 to 1%2F2%28log%28a%2C+++%28x-y%29%29-log%28a%2C%28x%2By%29%29%5E%22%22%29+?
Answer by Edwin McCravy(20065) About Me  (Show Source):
You can put this solution on YOUR website!
log%28a%2C+%28++%28x-y%29%2F%28sqrt%28x%5E2-y%5E2%29%29++%29%29 

Use this rule of logs:  red%28log%28P%2C%28Q%2FR%29%29=log%28P%2C%28Q%29%29-log%28P%2C%28R%29%29%29

log%28a%2C+++%28x-y%29%29-+log%28a%2C+%28sqrt%28x%5E2-y%5E2%29%29%29++

Now use this rule of changing radicals to fraction exponents:   and in particular  

log%28a%2C+++%28x-y%29%29-+log%28a%2C+%28x%5E2-y%5E2%29%5E%281%2F2%29%29++

Now use this rule of logs:  red%28++log%28P%2C%28Q%29%5ER%29+=+R%2Alog%28P%2C%28Q%29%29+%29

log%28a%2C+++%28x-y%29%29-+1%2F2log%28a%2C+%28x%5E2-y%5E2%29%29++

Let's put a 1 written as 2%2F2 in front of the first term, so
we can factor out a 1%2F2

2%2F2log%28a%2C+++%28x-y%29%29-+1%2F2log%28a%2C+%28x%5E2-y%5E2%29%29++

Factor out 1%2F2

1%2F2%282%2Alog%28a%2C+++%28x-y%29%29-+log%28a%2C+%28x%5E2-y%5E2%29%29%29++

Now let's factor x%5E2-y%5E2%29 as the difference of two squares, and get %28x-y%29%28x%2By%29.  So replacing that

1%2F2%282%2Alog%28a%2C+++%28x-y%29%29-+log%28a%2C+%28%28x-y%29%28x%2By%29%5E%22%22%29%29%29++

Use this rule of logs:  red%28log%28P%2C%28QR%29%29=log%28P%2C%28Q%29%29%2Blog%28P%2C%28R%29%29%29

1%2F2

Remove the inner parentheses which are preceded by a negative sign:

1%2F2

The first two log terms inside the big parentheses are like terms so
we can combine them:

1%2F2%28log%28a%2C+++%28x-y%29%29-log%28a%2C%28x%2By%29%29%5E%22%22%29+

Edwin