Answer
$x=\dfrac{1}{2}$
Work Step by Step
Using the properties of logarithms, the given equation, $
\log_3 x+\log_3 (2x+5)=1
$, is equivalent to
\begin{align*}\require{cancel}
\log_3 [x(2x+5)]&=1
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\
\log_3 (2x^2+5x)&=1
&(\text{use the Distributive Property})
.\end{align*}
Since $\log_b y=x$ implies $y=b^x$, the equation above is equivalent to
\begin{align*}\require{cancel}
2x^2+5x&=3^1
\\
2x^2+5x&=3
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
2x^2+5x-3&=0
.\end{align*}
Using the factoring of trinomials, the factored form of the equation above is
\begin{align*}\require{cancel}
(x+3)(2x-1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x+3=0 & 2x-1=0
\\
x=-3 & 2x=1
\\\\
& \dfrac{\cancel2x}{\cancel2}=\dfrac{1}{2}
\\\\
& x=\dfrac{1}{2}
.\end{array}
If $x=-3$, the term $\log_3 x$ of the original equation becomes $\log_3(-3)$. This is undefined since $\log_b x$ is defined only for $x$ and $b$ are positive real numbers.
Hence, the solution to the equation $
\log_3 x+\log_3 (2x+5)=1
$ is $
x=\dfrac{1}{2}
$.
