## Task B-3.1.

Solve the equation $\quad 9^{-(x+2)^{2}}+8 \cdot 3^{-x^{2}-4 x-5}-1=0$.
## Solution.

The given equation can be written in the form

$$
\begin{gathered}
9^{-(x+2)^{2}}+8 \cdot 3^{-\left(x^{2}+4 x+4\right)-1}-1=0 \\
9^{-(x+2)^{2}}+8 \cdot 3^{-(x+2)^{2}-1}-1=0 \\
9^{-(x+2)^{2}}+\frac{8}{3} \cdot 3^{-(x+2)^{2}}-1=0
\end{gathered}
$$

Let's introduce the substitution $3^{-(x+2)^{2}}=t$. This reduces the given equation to a quadratic equation.

$$
t^{2}+\frac{8}{3} t-1=0 \quad \text { or } \quad 3 t^{2}+8 t-3=0
$$

Its solutions are $t_{1}=-3, t_{2}=\frac{1}{3}$. Since we are dealing with an exponential function, the variable $t$ cannot be negative, so $t=\frac{1}{3}$, which implies $3^{-(x+2)^{2}}=3^{-1}$.

We get $-(x+2)^{2}=-1$

so $x+2=1$ or $x+2=-1$

and finally $x=-1$ or $x=-3$.