Solve the equation $\left(\frac{1+\sqrt{1-x^2}}{2}\right)^{\sqrt{1-x}} = (\sqrt{1-x})^{\sqrt{1-x}+\sqrt{1+x}}$

Question

Solve in $\mathbb{R}$: $ \left(\frac{1+\sqrt{1-x^2}}{2}\right)^{\sqrt{1-x}} = (\sqrt{1-x})^{\sqrt{1-x}+\sqrt{1+x}} $ My approach: Let $a = \sqrt{1-x}$ and $b = \sqrt{1+x}$ so $a^2 + b^2 = 2$. The equation becomes $\left(\frac{1+ab}{2}\right)^a = a^{a+b}$, which is equivalent to $\left(\frac{1+ab}{a^2+b^2}\right)^a = a^{a+b}$. After taking the natural logarithm, we get $a \ln(1+ab) - a \ln(a^2+b^2) = a \ln(a) + b \ln(a)$. I thought of considering a function but I couldn't find it. Any help is appreciated.

Answer 1

Clearly, $x = 0$ is a solution.

We claim that $x = 0$ is the only solution.

If $-1 \le x < 0$, we have $$\Big(\frac{1+\sqrt{1-x^2}}{2}\Big)^{\sqrt{1-x}} < 1 < (\sqrt{1-x})^{\sqrt{1-x}+\sqrt{1+x}}.$$

If $0 < x \le 1$, we have $$\Big(\frac{1+\sqrt{1-x^2}}{2}\Big)^{\sqrt{1-x}} \ge \frac{1+\sqrt{1-x^2}}{2} > \frac{\sqrt{1 - x} + \sqrt{1 - x}}{2},$$ and $$(\sqrt{1-x})^{\sqrt{1-x}+\sqrt{1+x}} \le \sqrt{1 - x}.$$ Thus, $\Big(\frac{1+\sqrt{1-x^2}}{2}\Big)^{\sqrt{1-x}} > (\sqrt{1-x})^{\sqrt{1-x}+\sqrt{1+x}}$.

The claim is proved.

Answer 2

I try follow yours ideas, but i first note $$1+ab = \frac{(a+b)^2}{2}$$

So we have $$(2a)In\left(\frac{a+b}{2}\right) = (a+b)In(a)$$

Equivalent to solve $In\left(\frac{a+b}{2}\right)=\left(\frac{a+b}{2a}\right)In(a)$.

This is basically $In(x)=\frac{x}{y}In(y)$

Which, I don’t know to solve, but wolfram solve for $a=b=1$. Hence $$\sqrt{1-x}=1 \\ \sqrt{1+x}=1$$

Has only $x=0$, which has the idea of the comment of @DMcMor. Sorry, this is incomplete if we don’t know solve $In(x)=\frac{y}{x}In(y)$, I read something about W lambert function.

Answer 3

Note that $a^2+b^2 = 2 \implies a^2+b^2+2ab = 2(1+ab)$. Hence $$\left(\frac{1+ab}{2}\right)^a = a^{a+b} \implies \left(\frac{a+b}{2}\right)^{2a} = a^{a+b} \implies \left(\frac{a+b}{2}\right)^{2/(a+b)} = a^{1/a} \tag{*}$$

Claim: $f(x) = x^{1/x}$ is injective for $x \in [0,e]$ (Define $f(0)$ as $\lim_{x \to 0}x^{1/x} = 0$).
Proof: We show it is strictly increasing in the said interval.
$f'(x) = x^{1/x} \left(\frac{1-\ln(x)}{x^2}\right)$ which is non-negative in the said interval (it is zero only at two points 0 and $e$, and otherwise strictly positive). $\blacksquare$

Using the claim on $(*)$, and noting that $a,\frac{a+b}2 \in [0,e]$ (as $a,b \le \sqrt{2}<e$) we must have $\frac{a+b}2 = a$, which gives the desired answer $\boxed{x=0}$.

Answer 4

COMMENT.-We have an equation with reals $a,b$ of the form $$\left(\frac{1+ab}{2}\right)^a=a^{a+b}$$ If I take $a=b$ then I get $$\left(\frac{1+a^2}{2}\right)^a=(a^2)^a\Rightarrow 1+a^2=2a^2\Rightarrow a^2=1$$ so $(\sqrt{1-x})^2=(\sqrt{1+x})^2=1$ and $x=0$ and the equation is verified. Thus $x=0$ is a solution.

To prove that it is the only solution is not immediate.I leave this here as a simple comment.