$7.38 x^{\log _{x^{2}}\left(x^{2}-1\right)}=5$.
$7.39\left\{\begin{array}{l}3^{2 x}-2^{y}=725 \\ 3^{x}-2^{\frac{y}{2}}=25\end{array}\right.$

$7.40\left\{\begin{array}{l}(x+y) 2^{y-2 x}=6.25, \\ (x+y)^{\frac{1}{2 x-y}}=5 .\end{array}\right.$

$7.41\left\{\begin{array}{l}2^{\frac{x-y}{2}}+2^{\frac{y-x}{2}}=2.5 \\ \lg (2 x-y)+1=\lg (y+2 x)+\lg 6\end{array}\right.$

$7.42\left\{\begin{array}{l}\log _{4} x-\log _{2} y=0, \\ x^{2}-2 y^{2}-8=0 .\end{array}\right.$

## Group 5

Solve the equations ( $7.43-7.50$ ):
7.38 According to the indication $2^{0}$,

$x>0$.

Further, taking into account the domain of the logarithmic function, we get $x^{2}-1>0$, or $|x|>1$, from which

$$
x>1
$$

From (1) and (2) it follows that $x>1$ (at the same time, the restriction $x \neq 1$ imposed on the base of the logarithm is automatically satisfied). Now we have

$$
x^{\log _{x} \sqrt{x^{2}-1}}=5 \text {, or } \sqrt{x^{2}-1}=5 ; x_{1,2}= \pm \sqrt{26} \text {. }
$$

The equation is satisfied only by $x=\sqrt{26}$.

Answer: $x=\sqrt{26}$.