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}

\)

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