Show that the equation $2(e^x - 3) = e^{\frac{1}{2}x}$ has only one solution and find this value correct to 3 significant figures.

1. Express the equation as a quadratic equation using a substitution.

Let $u = e^{\frac{1}{2}x}$. Then $e^x = (e^{\frac{1}{2}x})^2 = u^2$.
Substituting into the equation:
$2(u^2 - 3) = u$
$2u^2 - u - 6 = 0$

2. Solve the quadratic equation for $u$.

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

3. Show that there is only one valid solution by rejecting the negative root.

Substitute back $u = e^{\frac{1}{2}x}$:
$e^{\frac{1}{2}x} = -\frac{3}{2} \quad \text{or} \quad e^{\frac{1}{2}x} = 2$
Since $e^{\frac{1}{2}x} > 0$ for all real $x$, the equation $e^{\frac{1}{2}x} = -\frac{3}{2}$ has no real solution.
Therefore, there is only one solution, which comes from $e^{\frac{1}{2}x} = 2$.

4. Solve for $x$ and evaluate to 3 significant figures.

$e^{\frac{1}{2}x} = 2$
Take the natural logarithm on both sides:
$\frac{1}{2}x = \ln 2$
$x = 2\ln 2 \approx 1.38629$
$x = 1.39 \text{ (to 3 s.f.)}$