## Precalculus (6th Edition) Blitzer

$x\approx 0.69 \text{ or } x =0$
Let $u=e^x$ This means that: $e^{2x} = u^2$ Thus, the given equation can be written as: $u^2-3u+2=0$ Factor the trinomial. Look for factors of $2$ whose sum is $-3$. These factors are $-2$ and $-1$. Therefore the factored form of the trinomial is: $(u-2)(u-1)=0$ Use the zero product property by equating each factor to zero. $u-2 = 0 \text{ or } u-1=0$ Solve each equation to obtain: $u=2 \text{ or } u=1$ Switch $u$ back to $e^x$ to obtain: $e^x=2 \text{ or } e^x=1$ Solve each equation by taking the natural logarithm of both sides. $\ln{e^x} = 2 \text{ or } \ln{e^x}=1$ Use the rule $\ln{e^x} = x$ to obtain: $x = \ln{2} \text{ or } x=\ln{1}$ Use a calculator to solve for $x$. Round-off answers to two decimal places. $x\approx 0.69 \text{ or } x =0$