College Algebra (11th Edition)

$x\approx-0.123$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $2e^{5x+2}=8 ,$ use the properties of equality to isolate the $e$ expression. Then take the natural logarithm of both sides. Use the laws of logarithms and the properties of equality to isolate the variable. Finally, Express the answer with $3$ decimal places. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the $e$ expression results to \begin{array}{l}\require{cancel} \dfrac{2e^{5x+2}}{2}=\dfrac{8}{2} \\\\ e^{5x+2}=4 .\end{array} Taking the natural logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \ln e^{5x+2}=\ln4 .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} (5x+2)\ln e=\ln4 .\end{array} Since $\ln e =1,$ the equation above is equivalent to \begin{array}{l}\require{cancel} (5x+2)(1)=\ln4 \\\\ 5x+2=\ln4 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 5x=-2+\ln4 \\\\ x=\dfrac{-2+\ln4}{5} \\\\ x\approx-0.123 .\end{array}