Find the number of positive roots of equation $x^{x+1}=(x+1)^x$

Question

Find the number of positive roots of equation $$x^{x+1}=(x+1)^x$$

My work so far:

$$\ln x^{x+1}=\ln (x+1)^x$$ $$(x+1)\ln x=x\ln (x+1)$$ $$\frac{\ln x}{x}=\frac{\ln(x+1)}{x+1}$$ Let $f(t)=\frac{\ln t}{t}$. $$f'(t)=\frac{1-\ln t}{t^2}$$ Answer:1(one)

Answer 1

Let $g(x)=(x+1)\ln x-x\ln(x+1)$.

Then, $$g'(x)=\ln\frac{x}{x+1}+\frac{2x+1}{x(x+1)},\quad g''(x)=-\frac{x^2+x+1}{x^2(x+1)^2}$$

Since $g''(x)$ is negative, we see that $g'(x)$ is strictly decreasing with $\lim_{x\to\infty}g'(x)=0$ from which we have that $g'(x)$ is positive.

So, $g(x)$ is strictly increasing with $g(1)=-\ln 2\lt 0$ and $g(3)=\ln\frac{81}{64}\gt 0$.

Therefore, the number of positive solutions of $g(x)=0$ is $1$.

Answer 2

From your derivative, we see that $f$ has one maximum at $t=e$, and $f$ has one root, at $t=1$. For small $t$, $f(t)$ is very negative, while for large $t$, $f(t) \to 0$. We now notice that to solve $f(x)=f(x+1)$, we need $x<e$, $1+x>e$ (drawing the graph is an easy way to see this). There can only be one such point: imagine pushing a horizontal line segment of length $1$ along the graph of $f$, and you see that there is only one place where it touches $f$ twice. Alternatively, plot $f(t)$ and $f(1+t)$ and observe there's only one intersection.