Answer
$x=1$
Work Step by Step
Using the laws of logarithms, the given equation, $
\log(x+9)+\log x=1
,$ is equivalent to
\begin{array}{l}\require{cancel}
\log[(x+9)\cdot x]=1
\\\\
\log(x^2+9x)=1
.\end{array}
In exponential form, the logarithmic equation, $
\log(x^2+9x)=1
,$ is equivalent to
\begin{array}{l}\require{cancel}
\log_{10}(x^2+9x)=1
\\\\
x^2+9x=10^1
.\end{array}
Solving the equation, $
x^2+9x=10^1
,$ results in
\begin{array}{l}\require{cancel}
x^2+9x=10
\\\\
x^2+9x-10=0
\\\\
(x+10)(x-1)=0
\\\\
x=\{-10,1\}
.\end{array}
Upon checking only $
x=1
$ satisfies the original equation.
