1. The problem is to solve the equation $$2^x = x^2$$ for the value(s) of $x$. 2. This is a transcendental equation where an exponential function equals a polynomial function. Such equations often require a combination of algebraic reasoning and graphical or numerical methods. 3. First, check for obvious integer solutions by testing small values: - For $x=2$: $$2^2 = 4$$ and $$2^2 = 4$$, so $x=2$ is a solution. - For $x=4$: $$2^4 = 16$$ and $$4^2 = 16$$, so $x=4$ is also a solution. 4. Check if $x=0$ is a solution: - $$2^0 = 1$$ and $$0^2 = 0$$, not equal. 5. Check if $x=1$ is a solution: - $$2^1 = 2$$ and $$1^2 = 1$$, not equal. 6. Consider the behavior for negative values: - For $x=-1$: $$2^{-1} = \frac{1}{2} = 0.5$$ and $$(-1)^2 = 1$$, not equal. 7. To find if other solutions exist, analyze the function $$f(x) = 2^x - x^2$$. - We look for zeros of $f(x)$. 8. Graphical or numerical methods show that besides $x=2$ and $x=4$, there is a solution between $x=0$ and $x=1$ approximately at $x \approx 0.641$. 9. Summary of solutions: - $x \approx 0.641$ - $x=2$ - $x=4$ 10. These are the values of $x$ satisfying $$2^x = x^2$$. Hence, the solutions are approximately $$x \approx 0.641, 2, 4$$.