Certainly! Let's solve the equation [tex]\(2^x = x^2\)[/tex] step by step.
1. Identify the equation: We start with the equation [tex]\(2^x = x^2\)[/tex].
2. Consider potential solutions:
- Check for simple integer solutions:
- For [tex]\(x = 2\)[/tex]:
[tex]\[
2^2 = 4 \quad \text{and} \quad 2^2 = 4 \quad \Rightarrow \quad 4 = 4
\][/tex]
which is true. Therefore, [tex]\(x = 2\)[/tex] is a solution.
- For [tex]\(x = 4\)[/tex]:
[tex]\[
2^4 = 16 \quad \text{and} \quad 4^2 = 16 \quad \Rightarrow \quad 16 = 16
\][/tex]
which is true. Therefore, [tex]\(x = 4\)[/tex] is a solution.
- Check for other simpler negative or fractional solutions:
- For [tex]\(x = -1\)[/tex]:
[tex]\[
2^{-1} = \frac{1}{2} \quad \text{and} \quad (-1)^2 = 1 \quad \Rightarrow \quad \frac{1}{2} \neq 1
\][/tex]
- For [tex]\(x = 0\)[/tex]:
[tex]\[
2^0 = 1 \quad \text{and} \quad 0^2 = 0 \quad \Rightarrow \quad 1 \neq 0
\][/tex]
- Neither [tex]\(x = -1\)[/tex] nor [tex]\(x = 0\)[/tex] are solutions.
3. Identify the need for more complex solutions:
- Besides the simple integer values, there are complex and transcendental solutions which can be expressed in terms of the Lambert W function.
- For another solution, we have:
[tex]\[
x = -2 \cdot W\left(\frac{\log(2)}{2}\right) / \log(2)
\][/tex]
Here, [tex]\(W(z)\)[/tex] is the Lambert W function (also known as the product logarithm), which solves equations of the form [tex]\(z = W(z)e^{W(z)}\)[/tex].
4. Summary of all solutions:
- The solutions to the equation [tex]\(2^x = x^2\)[/tex] are:
[tex]\[
x = 2, \quad x = 4, \quad \text{and} \quad x = -2 \cdot W\left(\frac{\log(2)}{2}\right)/\log(2)
\][/tex]
Thus, these are the values of [tex]\(x\)[/tex] that satisfy the equation [tex]\(2^x = x^2\)[/tex].
