By using an appropriate substitution, solve the equation $e^{\frac{1}{2}x} = 2 + 24 e^{-\frac{1}{2}x}$.

1. Use a substitution to transform the equation into a quadratic form

Let $u = e^{\frac{1}{2}x}$. Then $e^{-\frac{1}{2}x} = \frac{1}{u}$.
$u = 2 + \frac{24}{u}$

2. Rearrange into a standard quadratic equation and solve for $u$

Multiply throughout by $u$:
$u^2 = 2u + 24$
$u^2 - 2u - 24 = 0$
$(u - 6)(u + 4) = 0$
$u = 6 \quad \text{or} \quad u = -4$

3. Substitute back and solve for $x$, rejecting invalid solutions

Since $e^{\frac{1}{2}x} > 0$ for all real $x$, we reject $u = -4$.
$e^{\frac{1}{2}x} = 6$
$\frac{1}{2}x = \ln 6$
$x = 2\ln 6 \approx 3.58$