[ASK] Logarithmic Equation

[Monox D. I-Fly]

A friend asked me how to solve this question:

[tex]log_2(x+2)+log_{(x-2)}4=3[/tex]

I said I had no idea because one is x + 2 and the other one is x - 2. If both are x + 2 or x - 2, I can do it. He said that if that's the case, even at his level he could solve it. This is what I've done so far regarding the question.

[tex]log_2(x+2)+log_{(x-2)}4=3[/tex]

[tex]\frac{log(x+2)}{log2}+\frac{log4}{log(x-2)}=3[/tex]

[tex]\frac{log(x+2)log(x-2)+log4log2}{log2log(x-2)}=3[/tex]

[tex]log(x+2)log(x-2)+2log^22=3log2log(x-2)[/tex]

[tex]log(x+2)log(x-2)-3log2log(x-2)=-2log^22[/tex]

[tex]log(x-2)(log(x+2)-3log2)=-2log^22[/tex]

What should I do from here? Or did I make some mistakes?

[HallsofIvy]

When you use the "conversion to another base" formula to write [tex]\frac{log(x+ 2)}{log(2)}+ \frac{log(4)}{log(x- 2)}= 3[/tex], the "log" can be to any base. Because of the "2" and "4", I would be inclined to use specifically base 2. In that case, the equation is [tex]log_2(x+ 2)+ \frac{2}{log_2(x- 2)}= 3[/tex].

[tex](log_2(x+ 2))(log_2(x- 2))+ 2= 3 log_2(x- 2)[//tex].

[tex](log_2(x+ 2))(log_2(x- 2))- 3 log_2(x- 2)= -2[/tex].

Let [tex]u= log_2(x+ 2)[/tex] and [tex]v= log_2(x- 2)[/tex].

Then the equation is uv- 3v= -2. Further, [tex]x+ 2= 2^u[/tex] and [tex]x- 2= 2^v[/tex] so that, subtracting, [tex]4= 2^u- 2^v[/tex].

So we have the two equations, [tex]uv- 3v= -2[/tex] and [tex]2^u- 2^v= 4[/tex] to solve for u and v. [tex](u- 3)v= -2[/tex] or [tex]u= \frac{3v- 2}{v}= 3- \frac{2}{v}[/tex] so [tex]2^{3- \frac{2}{v}- 2^v= \frac{8}{2^{2/v}- 2^v= 4[/tex].