## Task B-2.3.

Solve the equation in the set of real numbers

$$
4^{2 x+\sqrt{-1+x^{2}}}-5 \cdot 2^{2 x-1+\sqrt{-1+x^{2}}}=6
$$
## Solution.

We solve the problem by substitution $2 x+\sqrt{-1+x^{2}}=y$ under the condition $x^{2}-1 \geq 0$.

(Or by substitution $2^{2 x+\sqrt{-1+x^{2}}}=t$.)

Then we have $2 \cdot\left(2^{y}\right)^{2}-5 \cdot 2^{y}-12=0$.

The solution is only $2^{y}=4$ since $2^{y}>0 \forall y \in \mathbb{R}$. Then $y=2$.

It follows that $2 x+\sqrt{-1+x^{2}}=2$, or

$$
\sqrt{-1+x^{2}}=2-2 x
$$

After squaring, under the condition $2-2 x \geq 0$, we get

$$
3 x^{2}-8 x+5=0, \quad x_{1}=1, x_{2}=\frac{5}{3}
$$

Only the solution $x=1$ satisfies the given conditions.