https://file.aiquickdraw.com/mathbot/file/1725792475763k87sfari.jpg

The equation in the image is:

\[
(2^{-x})^x = x^y
\]

Let's solve for \(x\) and \(y\).

### Step 1: Simplify the left-hand side of the equation
\[
(2^{-x})^x = 2^{-x \cdot x} = 2^{-x^2}
\]
Thus, the equation becomes:
\[
2^{-x^2} = x^y
\]

### Step 2: Consider possible values of \(x\) and \(y\)
This equation is not straightforward to solve analytically without further context or constraints on \(x\) and \(y\). However, we can consider specific cases for \(x\) and \(y\) to see if they satisfy the equation.

#### Case 1: \(x = 1\)
\[
2^{-1^2} = 1^y \implies 2^{-1} = 1 \implies \frac{1}{2} = 1
\]
This is not true, so \(x = 1\) is not a solution.

#### Case 2: \(x = 2\)
\[
2^{-2^2} = 2^y \implies 2^{-4} = 2^y \implies 2^y = \frac{1}{16}
\]
Thus, \(y = -4\).

Therefore, one solution is \(x = 2\) and \(y = -4\).

### Conclusion
The solution to the equation

Find the values of x and y that satisfy the equation: (2^-x)^x = x^y

Learn how to solve the exponential equation (2^-x)^x = x^y step-by-step. This solution explores key concepts like exponent rules and algebraic manipulation, ideal for students in Grades 10-12.

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