4. Solve the inequality

$$
\frac{9^{x}-5 \cdot 15^{x}+4 \cdot 25^{x}}{-9^{x}+8 \cdot 15^{x}-15 \cdot 25^{x}}<0
$$
4. Let's write the inequality in the form

$$
\frac{\left(\frac{3}{5}\right)^{2 x}-5 \cdot\left(\frac{3}{5}\right)^{x}+4}{-\left(\frac{3}{5}\right)^{2 x}+8 \cdot\left(\frac{3}{5}\right)^{x}-15}<0
$$

5 points

By substituting $\left(\frac{3}{5}\right)^{x}=t$, we can rewrite the equation in a more convenient form

$$
\frac{t^{2}-5 t+4}{-t^{2}+8 t-15}<0
$$

or in factored form

$$
\frac{(t-1)(t-4)}{-(t-3)(t-5)}<0
$$

![](https://cdn.mathpix.com/cropped/2024_06_03_843fcc95a1719fe2319dg-07.jpg?height=339&width=439&top_left_y=1007&top_left_x=657)

The solution to this inequality is $t \in(0,1) \cup(3,4) \cup(5, \infty)$, or 10 points

$x \in\left(-\infty, \log _{\frac{3}{5}} 5\right) \cup\left(\log _{\frac{3}{5}} 4, \log _{\frac{3}{5}} 3\right) \cup(0, \infty)$.

10 points

## MINISTRY OF EDUCATION AND SPORT OF THE REPUBLIC OF CROATIA

CROATIAN MATHEMATICAL SOCIETY

## MATHEMATICS