## Task 21/69

The equation

$$
\sqrt[x]{16}+\sqrt[x]{20}=\sqrt[x]{25}
$$

is to be solved in the domain of real numbers.
After dividing both sides of the equation by $\sqrt[x]{25}$, we get

$$
\sqrt[x]{\frac{16}{25}}+\sqrt[x]{\frac{4}{5}}=1 \quad \text { or } \quad \sqrt[x]{\left(\frac{4}{5}\right)^{2}}+\sqrt[x]{\frac{4}{5}}=1
$$

It is natural to perform the substitution $a=\sqrt[x]{\frac{4}{5}}$. Then it follows that $a^{2}+a-1=0$ with $a=\frac{-1 \pm \sqrt{5}}{2}$.

The negative value is discarded; with the positive value, we get

$$
\sqrt[x]{\frac{4}{5}}=\frac{1+\sqrt{5}}{2}
$$

By taking the logarithm, we obtain the approximate solution $x=0.4637$. An estimation with $x \approx 0.5$ confirms the correctness of the result.