解方程 $2^{x+1} + 3(2^{-x}) = 7$。

1. Rewrite the equation using laws of indices to isolate $2^x$

$2(2^x) + \frac{3}{2^x} = 7$

2. Use a substitution to form a quadratic equation

Let $u = 2^x$.
$2u + \frac{3}{u} = 7$
$2u^2 - 7u + 3 = 0$

3. Solve the quadratic equation for $u$

$(2u - 1)(u - 3) = 0$
$u = \frac{1}{2} \text{ or } u = 3$

4. Solve for $x$ when $u = \frac{1}{2}$

$2^x = \frac{1}{2}$
$2^x = 2^{-1}$
$x = -1$

5. Solve for $x$ when $u = 3$

$2^x = 3$
$x \ln 2 = \ln 3$
$x = \frac{\ln 3}{\ln 2} \approx 1.5849 \approx 1.58 \text{ (to 3 s.f.)}$