Logarithms problems.

A friend asked me how to solve this question:

$$log_2(x+2)+log_{(x-2)}4=3$$

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.

$$log_2(x+2)+log_{(x-2)}4=3$$

$$\frac{log(x+2)}{log2}+\frac{log4}{log(x-2)}=3$$

$$\frac{log(x+2)log(x-2)+log4log2}{log2log(x-2)}=3$$

$$log(x+2)log(x-2)+2log^22=3log2log(x-2)$$

$$log(x+2)log(x-2)-3log2log(x-2)=-2log^22$$

$$log(x-2)(log(x+2)-3log2)=-2log^22$$

What should I do from here? Or did I make some mistakes?
Monox D. I-Fly

Posts: 18
Joined: Tue May 22, 2018 1:38 am
Reputation: 3

When you use the "conversion to another base" formula to write $$\frac{log(x+ 2)}{log(2)}+ \frac{log(4)}{log(x- 2)}= 3$$, 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 $$log_2(x+ 2)+ \frac{2}{log_2(x- 2)}= 3$$.

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

Let $$u= log_2(x+ 2)$$ and $$v= log_2(x- 2)$$.

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

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

HallsofIvy

Posts: 318
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 108