Prove that $(x+1)^x-x^x(x-1)$ only has one (real) root For $x>0$, I want to prove that $(x+1)^x-x^x(x-1)$ only has one root
 A: Write your equation as
$$ (1+1/x)^x = x-1 $$
Note that $x > 1$ is required for both sides to have the same sign, and then
$$ x \log(1+1/x) - \log(x-1) = 0$$
Call the left side $g(x)$.  Now show that


*

*$g(x)$ is convex.

*$\lim_{x \to 1+} g(x) = +\infty$

*$\lim_{x \to \infty} g(x) = -\infty$.


EDIT: As Clclstdnt comments, $g'' = \frac{x^3+x^2+3x-1}{x(x^2-1)^2}$.  For $x > 1$, both numerator and denominator are positive, so $g$ is convex on $(1,\infty)$.  Since $\lim_{x \to \infty} g(x) = -\infty$, this implies $g' < 0$ (i.e. if $g'(b) \ge 0$ for some $b$, we'd have $g'(x) \ge 0$ for $x \ge b$, and then $\lim_{x\to\infty} g(x)$ couldn't be $-\infty$).  That tells you $g$ has at most one zero.  On the other hand, (2) and (3) and the Intermediate Value Theorem say there is at least one.
A: My answer is definitely going to pull from some ideas of Robert Israel.
So first we re-write the equation as 
$$(1+1/x)^x = x-1$$ as previously suggested.  Then we consider these as two separate functions and ask when they intersect.
Let $f(x) = (1+1/x)^x$ and let $g(x)=x-1$. 
Motivated from the following answer How to prove $(1+1/x)^x$ is increasing when $x>0$?
we see that $\log(f(x))' = \log(1+\frac{1}{x})-\frac{1}{x+1}$ is increasing and hence $f(x)$ is too. Likewise one can easily show that $\log(f(x))'' = -\frac{1}{x(1+x)^2}<0$ this shows that $f$ is log-convex and hence convex (this follows from the following The composition of two convex functions is convex since $Exp$ is convex and $\log(f)$ is convex hence $f=Exp[\log(f)]$ is too).
$g(1)=0$ and $f(1)=2$ so $g<f$ at $x=1$. Now that we have established $f$ is convex it follows that once $g$ (a straight line) is greater than $f$ it is always greater. It follows there is one and only one solution for the equation equivalently $g$ intersects $f$ exactly once.
