## Precalculus (6th Edition) Blitzer

The value of x is $\left\{ \ln 2,\ln 4 \right\}$.
Consider the expression ${{e}^{2x}}-6{{e}^{x}}+8=0$ Simplify ${{e}^{2x}}$ ${{e}^{2x}}={{\left( {{e}^{x}} \right)}^{2}}$ Now, take ${{e}^{x}}=u$ So, the equation will be: ${{u}^{2}}-6u+8=0$ Factorize the above equation as follows: \begin{align} & {{u}^{2}}-6u+8=0 \\ & {{u}^{2}}-4u-2u+8=0 \\ & u\left( u-4 \right)-2\left( u-4 \right)=0 \\ & \left( u-4 \right)\left( u-2 \right)=0 \end{align} So, the factors are: \begin{align} & \left( u-4 \right)=0 \\ & u=4 \end{align} Or, \begin{align} & \left( u-2 \right)=0 \\ & u=2 \end{align} Now, substitute the value of $u=4$ in equation ${{e}^{x}}=u$: \begin{align} & {{e}^{x}}=u \\ & {{e}^{x}}=4 \\ \end{align} Take natural log on both sides; \begin{align} & \ln \left( {{e}^{x}} \right)=\ln \left( 4 \right) \\ & x=\ln \left( 4 \right) \end{align} Now, substitute the value of $u=2$ in equation ${{e}^{x}}=u$: \begin{align} & {{e}^{x}}=u \\ & {{e}^{x}}=2 \\ \end{align} Take natural log on both sides; \begin{align} & \ln \left( {{e}^{x}} \right)=\ln \left( 2 \right) \\ & x=\ln \left( 2 \right) \end{align} Hence, the value of x in the expression ${{e}^{2x}}-6{{e}^{x}}+8=0$ is $\left\{ \ln 2,\ln 4 \right\}$.