The diagram below shows the curve $y = 4e^{2x} - 7$ and the curve $y = 2e^{-2x}$.

1. Rearrange the given equation to match the equations of the curves in the diagram

The equation $4e^{2x} - 2e^{-2x} = 7$ can be rearranged as:
$4e^{2x} - 7 = 2e^{-2x}$

2. Relate the equation to the intersection of the curves and state the number of solutions

The solutions to $4e^{2x} - 7 = 2e^{-2x}$ correspond to the points of intersection of the curves $y = 4e^{2x} - 7$ and $y = 2e^{-2x}$.
From the diagram, the two curves intersect at exactly one point. Therefore, there is 1 solution.

1. Transform the equation into a quadratic equation in terms of $e^{2x}$

Let $u = e^{2x}$. Then $e^{-2x} = \frac{1}{u}$.
$4u - 7 = \frac{2}{u}$
$4u^2 - 7u - 2 = 0$

2. Solve the quadratic equation for $u$

$(4u + 1)(u - 2) = 0$
$u = -\frac{1}{4} \quad \text{or} \quad u = 2$

3. Substitute back $e^{2x}$ and reject the invalid solution

$e^{2x} = -\frac{1}{4} \quad \text{or} \quad e^{2x} = 2$
Since $e^{2x} > 0$ for all real $x$, $e^{2x} = -\frac{1}{4}$ has no real solution.

4. Solve for $x$ using natural logarithms

$e^{2x} = 2$
$2x = \ln 2$
$x = \frac{1}{2}\ln 2 \approx 0.347 \text{ (to 3 s.f.)}$