Solve the following equation:

$$
\left(5^{x}-2^{x-2}\right)^{2}+2 \log \left(5^{x}+2^{x-2}\right)=x
$$
Solution. Since $5^{x}>0$ and $2^{x-2}>0$ for all $x$, we can apply the arithmetic mean-geometric mean inequality to them:

$$
\frac{5^{x}+2^{x-2}}{2} \geq \sqrt{5^{x} \cdot 2^{x-2}}=\frac{\sqrt{10^{x}}}{2} .
$$

Multiply both sides by 2, then take the base-10 logarithm of both sides. Since the $\lg$ function is strictly increasing, this is equivalent to the following:

$$
\lg \left(5^{x}+2^{x-2}\right) \geq \lg 10^{\frac{x}{2}}=\frac{x}{2}
$$

Multiply both sides by 2, then add the inequality $\left(5^{x}-2^{x-2}\right)^{2} \geq 0$. This gives us:

$$
\left(5^{x}-2^{x-2}\right)^{2}+2 \lg \left(5^{x}+2^{x-2}\right) \geq x
$$

Equality can only hold if there is equality in the original inequalities, which in both cases happens when $5^{x}=2^{x-2}$, or $\left(\frac{2}{5}\right)^{x}=4$.

Take the base $\frac{2}{5}$ logarithm of both sides, then we get the only solution to the equation: $x=\log _{\frac{2}{5}} 4$.

If we simplify a bit more, we get the approximate value:

$$
x=\log _{\frac{2}{5}} 4=\frac{\lg 4}{\lg \frac{2}{5}} \approx-1.5129
$$