## Precalculus (6th Edition) Blitzer

$x\approx 1.10$
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. $x\approx 1.10$