Logs

[Guest]

I'm not sure how to do this one and would appreciate any help.
(Logx)^2-(Logy)^2

The answer I'm supposed to get is (Logxy)(Log(x/y)), but I don't know how to get there.

 

[soroban]

Hello, BW52!

(log x)² - (log y)²

The answer is: [log(xy)] [log(x/y)]

We have a "difference of squares" which factors:

. . . (log x)² - (log y)² . = . (log x + log y)(log x - log y)

Get it?

 

[Guest]

I see, now.
So from there:
(Logx+Logy)(Logx-Logy)
combine the Logs
(Logxy)(Log(x/y))

Now it works! Thank you!