## College Algebra (11th Edition)

$x=6$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\ln e^{\ln x}-\ln(x-4)=\ln3 ,$ use the laws of logarithms. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ 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} (\ln x)(\ln e)-\ln(x-4)=\ln3 .\end{array} Since $\ln e=1,$ the equation above is equivalent to \begin{array}{l}\require{cancel} (\ln x)(1)-\ln(x-4)=\ln3 \\\\ \ln x-\ln(x-4)=\ln3 .\end{array} Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \ln \dfrac{x}{x-4}=\ln3 .\end{array} Since both sides have the same logarithmic base, then the logarithm can be dropped. That is, \begin{array}{l}\require{cancel} \dfrac{x}{x-4}=3 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} (x-4)\left( \dfrac{x}{x-4} \right)=(x-4)(3) \\\\ x=3x-12 \\\\ x-3x=-12 \\\\ -2x=-12 \\\\ x=\dfrac{-12}{-2} \\\\ x=6 .\end{array}