We are asked to solve the equation $2^{\sqrt{2x+1}} = 32$ for $x$.

1. Problem Description

We are asked to solve the equation $2^{\sqrt{2x+1}} = 32$ for $x$.

2. Solution Steps

The given equation is $2^{\sqrt{2x+1}} = 32$. We can rewrite 32 as $2^5$. Thus, the equation becomes

$2^{\sqrt{2x+1}} = 2^5$.

Since the bases are the same, we can equate the exponents:

$\sqrt{2x+1} = 5$.

Squaring both sides, we get:

$(\sqrt{2x+1})^2 = 5^2$, which simplifies to

$2x+1 = 25$.

Subtracting 1 from both sides, we have:

$2x = 25 - 1$, so

$2x = 24$.

Dividing both sides by 2, we get:

$x = \frac{24}{2}$, which simplifies to

$x = 12$.

3. Final Answer

The solution to the equation is $x = 12$. So the answer is C.

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