Answer
$x=1$
Work Step by Step
Using the properties of logarithms, the given equation, $
\log (2x-1)+\log 10x=\log10
$, is equivalent to
\begin{align*}\require{cancel}
\log [(2x-1)(10x)]&=\log10
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\
\log (20x^2-10x)&=\log10
&(\text{use the Distributive Property})
.\end{align*}
Since $\log_b x=\log_b y$ implies $x=y$, then the equation above implies
\begin{align*}\require{cancel}
20x^2-10x&=10
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
20x^2-10x-10&=0
\\\\
\dfrac{20x^2-10x-10}{10}&=\dfrac{0}{10}
\\\\
2x^2-x-1&=0
.\end{align*}
Using the factoring of trinomials, the factored form of the equation above is
\begin{align*}\require{cancel}
(x-1)(2x+1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x-1=0 & 2x+1=0
\\
x=1 & 2x=-1
\\\\
& \dfrac{\cancel2x}{\cancel2}=\dfrac{-1}{2}
\\\\
& x=-\dfrac{1}{2}
.\end{array}
If $x=-\dfrac{1}{2}$, the term $\log(2x-1)$ of the original equation becomes $\log(-2)$. 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 (2x-1)+\log 10x=\log10
$ is $
x=1
$.
