## Task 19/82

Solve the equation $25^{x}-30^{x}=36^{x+0.5}$.
Due to

$$
25^{x}-30^{x}=36^{x+0.5}=36^{x} \sqrt{36}=6 \cdot 36^{x}
$$

it is reasonable to first divide the equation by $36^{x} \neq 0$. This yields

$$
\frac{25^{x}}{36^{x}}-\frac{30^{x}}{36^{x}}=\left[\left(\frac{5}{6}\right)^{2}\right]^{x}-\left(\frac{5}{6}\right)^{x}=6
$$

and with the substitution $\left(\frac{5}{6}\right)^{x}=t$ we obtain the quadratic equation in $t$: $t^{2}-t-6=0$ with the solutions $t_{1 ; 2}=0.5 \pm 2.5$, of which only $t_{1}=3$ is valid due to $t>0$. From this, it follows that

$$
x=\log _{z} 3=\frac{\lg 3}{\lg z}=\frac{\lg 3}{\lg 5-\lg 6} \approx-6.03
$$

with $z=\frac{5}{6}$.