## Task 14/75

We are looking for all solutions of the equation $a^{2 x}+a^{x} b^{x}=b^{2 x}$, where $a$ and $b$ are positive real numbers.

What additional condition must $a$ and $b$ satisfy?
From $a^{2 x}+a^{x} b^{x}=b^{2 x}$, due to $a \neq 0$, by dividing with $a^{2 x}$ and after equivalent transformation,

$$
\frac{b^{2 x}}{a^{2 x}}-\frac{b^{x}}{a^{x}}-1=0 \quad ; \quad\left[\left(\frac{b}{a}\right)^{x}\right]^{2}-\left(\frac{b}{a}\right)^{x}-1=0
$$

By setting $\left(\frac{b}{a}\right)^{x}=z$, the equation takes the form $z^{2}-z-1=0$ with the solution

$$
z=\left(\frac{b}{a}\right)^{x}=\frac{1+\sqrt{5}}{2}
$$

(due to $z>0$, the second solution is discarded). From this, by taking the logarithm,

$$
x \ln \frac{b}{a}=\ln \frac{1+\sqrt{5}}{2} \quad ; \quad x=\frac{\ln (1+\sqrt{5})-\ln 2}{\ln b-\ln a}
$$

Thus, there exists exactly one solution under the condition that $a \neq b$ (otherwise the denominator would be zero).