#### Answer

$2;$ $\dfrac {-5}{2}$

#### Work Step by Step

$\log x+\log \left( 2x+1\right) =1\Rightarrow \log _{10}\left[ x\left( 2x+1\right) \right] =1$
$\Rightarrow 2x^{2}+x=10^{1}\Rightarrow 2x^{2}+x-10=0$
$\Rightarrow x_{1},2=\dfrac {-b\pm \sqrt {b^{2}-4ac}}{2a}=\dfrac {-1\pm \sqrt {1^{2}-4\times 2\times \left( -10\right) }}{2\times 2}=\dfrac {-1\pm 9}{4}=2;\dfrac {-5}{2}$