## Algebra 2 Common Core

$x=\pm \sqrt{e^{4.9}} \approx \pm 11.588$
Subtract $1.1$ from each side to obtain: \begin{align*} 1.1+\ln{x^2}-1.1&=6-1.1\\ \ln{x^2}&=4.9 \end{align*} Recall: $$\ln{a} =y \longleftrightarrow e^y=a$$ Use the rule above to obtain: \begin{align*} \ln{x^2}&=4.9\\ e^{4.9}&=x^2 \end{align*} Take the square root of both sides: \begin{align*} \pm\sqrt{e^{4.9}}&=x\\ \end{align*} Use a calculator to obtain: $$x\approx \pm11.588$$