# Logs/Calc

#### A Beautiful Mind

Let $$\displaystyle f(x) = log \frac{1+x}{1-x}$$ if $$\displaystyle x > 0$$. If $$\displaystyle a$$ and $$\displaystyle b$$ are given numbers, with $$\displaystyle ab$$ not equal to $$\displaystyle -1$$, find all $$\displaystyle x$$ such that $$\displaystyle f(x) = f(a)+f(b)$$.

I've thought of a couple things to do.

Of course I thought about switching it from $$\displaystyle log (1+x) - log (1-x)$$. And then I thought I'd take the integral of both evaluated from a to b, but that turned out to be a mess. Any hints/suggestions or even a solution?

#### skeeter

MHF Helper
Let $$\displaystyle f(x) = log \frac{1+x}{1-x}$$ if $$\displaystyle x > 0$$. If $$\displaystyle a$$ and $$\displaystyle b$$ are given numbers, with $$\displaystyle ab$$ not equal to $$\displaystyle -1$$, find all $$\displaystyle x$$ such that $$\displaystyle f(x) = f(a)+f(b)$$.

I've thought of a couple things to do.

Of course I thought about switching it from $$\displaystyle log (1+x) - log (1-x)$$. And then I thought I'd take the integral of both evaluated from a to b, but that turned out to be a mess. Any hints/suggestions or even a solution?
no calculus required here ... more of a pain-in-the-@ algebra drill.

$$\displaystyle f(x) = \log \left(\frac{1+x}{1-x}\right)$$

if $$\displaystyle x > 0$$ , the the domain of $$\displaystyle f(x)$$ is $$\displaystyle 0 < x < 1$$

$$\displaystyle f(x) = f(a) + f(b)$$

$$\displaystyle \log \left(\frac{1+x}{1-x}\right) = \log \left(\frac{1+a}{1-a}\right) + \log \left(\frac{1+b}{1-b}\right)$$

$$\displaystyle \log \left(\frac{1+x}{1-x}\right) = \log \left(\frac{1+a}{1-a} \cdot \frac{1+b}{1-b}\right)$$

$$\displaystyle \frac{1+x}{1-x} = \frac{1+a}{1-a} \cdot \frac{1+b}{1-b}$$

cross-multiply ...

$$\displaystyle (1+x)(1-a)(1-b) = (1-x)(1+a)(1+b)$$

distribute $$\displaystyle (1+x)$$ and $$\displaystyle (1-x)$$ ...

$$\displaystyle (1-a)(1-b) + x(1-a)(1-b) = (1+a)(1+b) - x(1+a)(1+b)$$

$$\displaystyle x(1+a)(1+b) + x(1-a)(1-b) = (1+a)(1+b) - (1-a)(1-b)$$

$$\displaystyle x[(1+a)(1+b) + (1-a)(1-b)] = (1+a)(1+b) - (1-a)(1-b)$$

$$\displaystyle x = \frac{(1+a)(1+b) - (1-a)(1-b)}{(1+a)(1+b) + (1-a)(1-b)}$$

$$\displaystyle x = \frac{1+b+a+ab - (1-a-b+ab)}{1+b+a+ab + 1-a-b+ab}$$

$$\displaystyle x = \frac{2(a+b)}{2(1+ab)} = \frac{a+b}{1+ab}$$