Logarithmic equation: $\log_2 (\frac{x+1}{x^2+2} )= x^2-2x-1$

Question

Find every real numbers x so that $\log_2 (\frac{x+1}{x^2+2} )= x^2-2x-1$

Of course $x=${$0$,$2$} are solutions. Because $x^2-2x-1$ is strictly increasing from $[1,\infty )$ and $\log_2 (\frac{x+1}{x^2+2} )$ is decreasing in that interval ,$2$, is the only solution in $[1,\infty )$. How can i prove that $0$ is the only one remaining?

Answer 1

$$\eqalign{ & {\log _2}\left( {\frac{{x + 1}}{{{x^2} + 2}}} \right) = {x^2} - 2x - 1 \cr & * x > - 1 \cr & \Leftrightarrow 2\left( {x + 1} \right) + {\log _2}\left( {2\left( {x + 1} \right)} \right) = {x^2} + 2 + {\log _2}\left( {{x^2} + 2} \right) \cr & {\text{Let}}:f\left( t \right) = t + {\log _2}\left( t \right),t > 0 \cr & f'\left( t \right) = 1 + \frac{1}{{t\ln \left( 2 \right)}} > 0\forall t > 0 \Rightarrow f\left( t \right){\text{ is strictly increasing}} \cr & \Rightarrow f\left( {2\left( {x + 1} \right)} \right) = f\left( {{x^2} + 2} \right) \Leftrightarrow {x^2} + 2 = 2x + 2 \Leftrightarrow {x^2} - 2x = 0 \cr & \Rightarrow x = 0,x = 2 \cr} $$

Answer 2

When $x$ is in $(0,2)$, $\frac{x+1}{x^2+2}$ is in $(1/2,1)$. Indeed, $\frac{x+1}{x^2+2}>1/2$ is equivalent to $2x+2>x^2+2$, or $2>x$, which is true. And $\frac{x+1}{x^2+2}<1$ comes from $x+1<x^2+2$, or $x^2-x+1>0$, which is always true.

Then $\log_2\frac{x+1}{x^2+2}$ is in $(-1,0)$ for $x\in (0,2)$.

Also, for $x\in (0,2)$ holds $x^2-2x-1< -1$, or $x^2-2x<0$, or $x(x-2)<0$, which is true.

So the graphs of the LHS and the RHS are separated by a horizontal line $y=-1$.

If $x<0$, then $\frac{x+1}{x^2+2} <1/2$, since $2x+2<x^2+2$, or $0<x(x-2)$, which is true. Hence $\log_2\frac{x+1}{x^2+2}<-1$.

And $x^2-2x-1> -1$, since $x(x-2)>0$, which is true.

So the graphs of the LHS and the RHS are separated by a horizontal line $y=-1$ again, although they changed half-planes.

Therefore, there are no roots when $x\in (-\infty,2)\setminus\{0\}$.

Here are the graphs, for reference: WolframAlpha