Solve $\sqrt[x]{\frac{2}{(x+1)}} = {(x+1)}^{x+2}$.

Question

Please, help me solve the following exponential equation:

$$\sqrt[x]{\frac{2}{(x+1)}} = {(x+1)}^{x+2}$$

for $x \in \Bbb{R}$

P.S.: Sorry for all the formatting issues, I'm new in the site.

Answer 1

Let $t=(x+1)^2$ to recast the equation

$$\sqrt[x]{\frac{2}{(x+1)}} = {(x+1)}^{x+2}$$

as

$$t\ln t=2\ln2$$

which yields $t=2$, and in turn leads to the solution $x=\sqrt2-1$.

Note that $t\ln t \le 0 $ over $(0,1]$ and $(t\ln t)’>0$ for $t>1$, a strictly increasing function to assure one unique root.

Answer 2

Straightforward manipulations can transform the problem into solving $$ (x+1)^{x^2+2x+1} = 2. $$ The left-hand side is at most $1$ or undefined for $x\le0$. For $x>0$, it's not hard to show that the derivative is always positive, so that the left-hand side is increasing there and so any solution must be unique; furthermore, the values at $x=0$ and $x=1$, namely $1$ and $16$, show that a solution must exist between those two values.

I don't know a way to algebraically solve that equation, but finding an approximate root led me to conjecture that $x=\sqrt2-1$ is the solution, which (once written down) is easy to verify.