#### Answer

$x=2$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\log x+\log(2x+1)=1
,$ use the properties of logarithms to simplify the left-hand expression. Then change to exponential form. Express the resulting equation in the form $ax^2+bx+c=0$ and use the Quadratic Formula to solve for the values of the variable. Then do checking of the solutions with the original equation.
$\bf{\text{Solution Details:}}$
Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log [x(2x+1)]=1 .\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} [x(2x+1)]=1 \\\\ x(2x+1)=10^1 \\\\ 2x^2+x=10 \\\\ 2x^2+x-10=0 .\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is
\begin{array}{l}\require{cancel}
(2x+5)(x-2)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then the solutions are
\begin{array}{l}\require{cancel}
2x+5=0
\\\\\text{OR}\\\\
x-2=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
2x+5=0
\\\\
2x=-5
\\\\
x=-\dfrac{5}{2}
\\\\\text{OR}\\\\
x-2=0
\\\\
x=2
.\end{array}
If $
x=-\dfrac{5}{2}
,$ the part of the given expression, $
\log x
,$ becomes $
\log \left(-\dfrac{5}{2} \right)
.$ This is not allowed since $
\log x$ is defined only for positive values of $x.$ Hence, only $
x=2
$ satisfies the original equation.