College Algebra (11th Edition)

$\text{no solution}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\ln x-4\ln 3=\ln\left( \dfrac{1}{5} x \right) ,$ use the properties of logarithms to simplify the expression at the left. Then drop the logarithm on both sides and 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 3^4=\ln\left( \dfrac{1}{5} x \right) \\\\ \ln x-\ln 81=\ln\left( \dfrac{1}{5} x \right) .\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}{81}=\ln\left( \dfrac{1}{5} x \right) .\end{array} Since both sides use the same logarithmic base, then the logarithm can be dropped. Hence, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{x}{81}=\dfrac{x}{5} .\end{array} Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to \begin{array}{l}\require{cancel} x(5)=81(x) \\\\ 5x=81x \\\\ 5x-81x=0 \\\\ -76x=0 \\\\ x=0 .\end{array} If $x=0,$ the part of the given equation, $\ln x ,$ becomes $\ln 0$ which is not a real number. Hence, there is $\text{ no solution .}$