Answer
$x=\dfrac{2}{3}$
Work Step by Step
$\log_{2}x+\log_{2}(3x+1)=1$
Combine $\log_{2}x+\log_{2}(3x+1)$ as the $\log$ of a product:
$\log_{2}x(3x+1)=1$
$\log_{2}(3x^{2}+x)=1$
Rewrite in exponential form:
$2^{1}=3x^{2}+x$
$3x^{2}+x=2$
Take the $2$ to the left side of the equation:
$3x^{2}+x-2=0$
Solve this equation by factoring:
$(x+1)(3x-2)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x+1=0$
$x=-1$
$3x-2=0$
$3x=2$
$x=\dfrac{2}{3}$
The original equation is undefined for $x=-1$, so the answer is just $x=\dfrac{2}{3}$
