Logarithms

[FNongue]

Please I need help.

[Guest]

Your equation is [tex]2^{f(x- 2)}\times 2^{f(2x+2)}= 4^{log(f(x)}[/tex]
Since [tex]4= 2^2[/tex], that is the same as [tex]2^{f(x- 2)}\times 2^{f(2x+2)}= 2^{2log(f(x)}[/tex].

Taking the logarithm base 2 that becomes [tex]f(x-2)+ f(2x+2)= 2 log(f(x))[/tex].

We are told that f(x)= log(x+ 2) so f(x+ 2)= log((x+2)+2)= log(x+ 4), f(2x+ 2)= log((2x+2)+ 2)= log(2x+ 4). and log(f(x))= log(log(x+ 2)). So [tex]log(x+ 4)+ log(2x+2)= log((x+4)(2x+2))= 2 log(log(x+2))= log((log(x+2)) ^2). Then (x+4)(2x+2)= [log(x+2)] ^ 2.

That's about as far as I can go! Frankly, I would recommend that you re-read the problem. It is that "log(f(x))" that is causing trouble.