TITLE: An exponential equation
QUESTION [5 upvotes]: Need to solve the equation $$(x+1)^{x-1}=(x-1)^{x+1}$$ After applying logarithm on each side one obtains the following equation: $$f(x+1)=f(x-1)\text{, where }f(x)=\ln x/x $$
which doesn't seem to have a solution judging from the graph of $f$. What approach would you suggest? Thank you!

REPLY [2 votes]: Let's work with
$$
(x-1)\ln(x+1)=(x+1)\ln(x-1)
$$
and $x>1$.  As my comment states, $x=3$ solves this.  Consider
$$
g(x)=(x-1)\ln(x+1)-(x+1)\ln(x-1).
$$
Then
$$
g'(x)=\ln\left(\frac{x+1}{x-1}\right)-\frac{4x}{x^2-1}.
$$
Note that $\lim_{x\to \infty}g'(x)=0$.  Also note that
$$
g''(x)=\frac{2x^2+6}{(x^2-1)^2}
$$
which is positive for $x>1$.  Thus $g'(x)$ is an increasing function that approaches $0$.  Thus $g'(x)<0$ for all $x>1$.  Therefore $g(x)$ is decreasing for all $x>1$.