## Task A-3.1.

For which real numbers $x$ does the following hold

$$
5^{2 x}+4^{x}<29 \cdot 10^{x-1} ?
$$
## Solution.

Let's divide the initial inequality by $10^{x}$ and rearrange the expression on the left side of the inequality. Since $10^{x}$ is always a positive number, the direction of the inequality does not change. We proceed as follows:

$$
\begin{aligned}
\frac{5^{2 x}}{10^{x}} + \frac{4^{x}}{10^{x}} &< \frac{29 \cdot 10^{x}}{10^{x}} \\
\left(\frac{5}{2}\right)^{2 x} + \left(\frac{2}{5}\right)^{x} &< 29 \\
10 t^{2} + t &< 29 \quad \text{where } t = \left(\frac{5}{2}\right)^{x} \text{ and } t > 0.
\end{aligned}
$$

We obtain the quadratic inequality $10 t^{2} - 29 t + 10 < 0$.

The solution to the quadratic inequality is $t \in \left\langle \frac{2}{5}, \frac{5}{2} \right\rangle$.

From the substitution $t = \left(\frac{5}{2}\right)^{x}$, we get $\left(\frac{5}{2}\right)^{x} \in \left\langle \frac{2}{5}, \frac{5}{2} \right\rangle$, or equivalently,

$$
\frac{2}{5} < \left(\frac{5}{2}\right)^{x} < \frac{5}{2}
$$

Applying $\log_{\frac{5}{2}}$ to the above inequalities, we get $-1 < x < 1$ (since the base of the logarithm is greater than 1, the inequality does not change), from which we obtain the final solution $x \in \langle -1, 1 \rangle$.

Note: The solution could also be carried out by dividing the initial inequality by $4^{x}$ or $5^{2 x}$.

If students do not comment that the inequality does not change when applying a logarithm with a positive base or multiplying by a positive number that depends on the variable, they lose 1 point for each such inequality, with a maximum deduction of 2 points.