Solve exponential equation under some conditions

Question

Let $a, b, c$ be strictly positive real numbers. Solve the equation $2a^x = b^x + c^x$, given that $c = \sqrt{ab}$.

Knowing that $c = \sqrt{ab}$, we get $c^x = \frac{1}{a^xb^x}$. By plugging this in the equation, we get $a^xb^x(2a^x - b^x) = 1$.

Knowing that $c^2 = ab$, we get $c^{2x} = a^xb^x$. By plugging this in previously obtained equation, we get $c^{2x}(2b^x + c^x) = 1$.

That's all I could find! I have no idea how to proceed.

Thank you in advance!

Answer 1

You can devide your equation by $\sqrt{ab}^x$ and e.g. set $z:=(\frac{a}{b})^{\frac{x}{2}}$.

Then you get $2z=\frac{1}{z}+1$ which can be solved easily:

$z\in\{-\frac{1}{2};1\}$ and therefore $x=0$ is the only solution for $a\ne b$. The case $a=b$ is trivial.

Answer 2

HINT:

What if $ab=0$

Else let $$\sqrt{\dfrac cb}=\sqrt{\dfrac ba}=t\implies b=at^2,c=\cdots=at^4$$

$$\implies2a^z=(at^2)^z+(at^4)^z\implies(t^{2z})^2+(t^{2z})-2=0$$

Now $(t^{2z})^2+(t^{2z})-2=(t^{2z}+2)(t^{2z}-1)$