#### Answer

$x=-1$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
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}