Sketch the graph of $y = e^x + 2$. Solve the equation $3 - e^{-x} = 2e^x$.

1. Identify the horizontal asymptote and the general shape of the curve

The base graph is $y = e^x$, which has a horizontal asymptote at $y = 0$ and increases exponentially.
The graph of $y = e^x + 2$ is a vertical translation of $y = e^x$ upwards by 2 units.
Therefore, the horizontal asymptote is $y = 2$.

2. Find the $y$-intercept and sketch the graph

When $x = 0$, $y = e^0 + 2 = 1 + 2 = 3$.
The $y$-intercept is $(0, 3)$.
Sketch the curve passing through $(0, 3)$, approaching the line $y = 2$ as $x \to -\infty$, and increasing as $x \to \infty$.

1. Rewrite the equation as a quadratic equation in terms of $e^x$.

Let $u = e^x$. Then $e^{-x} = \frac{1}{u}$.
$3 - \frac{1}{u} = 2u$
Multiply by $u$:
$3u - 1 = 2u^2$
$2u^2 - 3u + 1 = 0$

2. Solve the quadratic equation for $u$.

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

3. Solve for $x$ by taking the natural logarithm.

Substitute back $u = e^x$:
$e^x = \frac{1}{2} \implies x = \ln\left(\frac{1}{2}\right) = -\ln 2$
$e^x = 1 \implies x = \ln 1 = 0$