Question

How do you condense `2log(x-3)+log(x+2)-6logx`?

Answer 1

`2log(x-3)+log(x+2)-6logx` `=log(((x-3)^2 (x+2))/x^6)`

Explanation:

`2log(x-3)+log(x+2)-6logx`
`=log(x-3)^2+log(x+2)-logx^6`-> use property `log_bx^n=nlog_bx`
`=log(((x-3)^2 (x+2))/x^6)`->use properties `log_b(xy)=log_bx+log_by, and log_b(x/y) = log_b x-log_b y`