#### Answer

$x\approx 1.10$

#### Work Step by Step

Let
$u=e^x$
This means that:
$e^{2x} = u^2$
Thus, the given equation can be written as:
$u^2-2u-3=0$
Factor the trinomial. Look for factors of $-3$ whose sum is $-2$.
These factors are $-3$ and $1$.
Therefore the factored form of the trinomial is:
$(u-3)(u+1)=0$
Use the zero product property by equating each factor to zero.
$u-3 = 0 \text{ or } u+1=0$
Solve each equation to obtain:
$u=3 \text{ or } u=-1$
Switch $u$ back to $e^x$ to obtain:
$e^x=3 \text{ or } e^x=-1$
Note that the second equation has no solution since the value of $e^x$ is always non-negative.
Solve the first equation by taking the natural logarithm of both sides.
$\ln{e^x} = \ln3$
Use the rule $\ln{e^x} = x$ to obtain:
$x = \ln{3}$
Use a calculator to solve for $x$. Round-off answers to two decimal places.
$x\approx 1.10$