## College Algebra (11th Edition)

$x=-1$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\log(3x+5)-\log(2x+4)=0 ,$ use the properties of logarithms to simplify the left-hand expression. Then change to exponential form. Use the properties of equality to isolate the variable. Finally, do checking of the solution with the original equation. $\bf{\text{Solution Details:}}$ Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log\dfrac{3x+5}{2x+4}=0 .\end{array} Since $y=b^x$ is equivalent to $\log_b y=x,$ the exponential form of the equation above is \begin{array}{l}\require{cancel} \log_{10}\dfrac{3x+5}{2x+4}=0 \\\\ \dfrac{3x+5}{2x+4}=10^0 \\\\ \dfrac{3x+5}{2x+4}=1 .\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} (3x+5)(1)=(2x+4)(1) \\\\ 3x+5=2x+4 \\\\ 3x-2x=4-5 \\\\ x=-1 .\end{array}