## College Algebra (11th Edition)

$x=\pm e^{3}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\ln x+\ln x^3=12 ,$ use the properties of logarithms to simplify the left side. Then use the definition of the natural logarithm and convert to exponential form. Finally, use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \ln (x\cdot x^3)=12 \\\\ \ln x^4=12 .\end{array} Since $\ln x=\log_e x,$ the equation above is equivalent to \begin{array}{l}\require{cancel} \log_e x^4=12 .\end{array} Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} x^4=e^{12} .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} \left(x^4\right)^{1/4}=\pm\left(e^{12}\right)^{1/4} \\\\ x=\pm e^{\frac{12}{4}} \\\\ x=\pm e^{3} .\end{array}