## Task B-3.1.

Solve the equation $5 \cdot 0.16^{x}+8 \cdot 0.4^{x}=4$.
## Solution.

Let $t=(0.4)^{x}$. 1 point

After introducing the substitution, the equation takes the form $5 t^{2}+8 t-4=0$.

The solutions of the obtained quadratic equation are:

$t_{1,2}=\frac{-8 \pm \sqrt{64+80}}{10}=\frac{-8 \pm \sqrt{144}}{10}=\frac{-8 \pm 12}{10}$,

which means $t_{1}=\frac{2}{5}$ and $t_{2}=-2$.

It remains to determine the values of the unknown $x$.

From the equation $0.4^{x}=\frac{2}{5}$, we get $\left(\frac{2}{5}\right)^{x}=\frac{2}{5}$ and the solution is $x=1$.

There is no real number $x$ for which $0.4^{x}=-2$.

The solution $x=1$ is the only solution to the original equation.