11.5. Solve the system of equations $\left\{\begin{array}{l}3^{x}=\sqrt{y} \\ 2^{-y}=x^{3}\end{array}\right.$.
Answer. $x=1 / 2, y=3$. Solution. It is directly verified that the numbers $x=1 / 2, y=3$ are a solution to the system. We will prove that the solution is unique. For this, we will show that the function $y(x)$, defined by the first equation, is strictly decreasing, and the function defined by the second equation is strictly decreasing. Indeed, the derivative of the first function is $y'=\left(3^{2 x}\right)'=\ln 3 \cdot 3^{2 x} \cdot 2>0$. The second function is defined for $x>0$ and its derivative has the form $y'=\left(-\frac{3 \ln x}{\ln 2}\right)'=-\frac{3}{\ln 2} \cdot \frac{1}{x}<0$. Thus, the system has a unique solution $x=1 / 2, y=3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ca5361dea44456c8863cg-1.jpg?height=497&width=800&top_left_y=2133&top_left_x=1162)

When justifying the monotonicity of the specified functions, one can also avoid using the derivative and refer to the corresponding properties of exponential and logarithmic functions for specific bases.