Answer
$\left\{\dfrac{5}{2}\right\}$
Work Step by Step
We are given the equation:
$\log_3 (x-1)+\log_3 (2x+1)=2$
Use the Product Property of logarithms:
$\log_3 (x-1)(2x+1)=2$
Rewrite the equation in exponential form:
$(x-1)(2x+1)=3^2$
$2x^2+x-2x-1-9=0$
$2x^2-x-10=0$
Solve for $x$:
$x=\dfrac{1\pm\sqrt{(-1)^2-4(2)(-10)}}{2(2)}=\dfrac{1\pm 9}{4}$
$x_1=\dfrac{1-9}{4}=-2$
$x_2=\dfrac{1+9}{4}=\dfrac{5}{2}$
Check the solutions:
$x_1=-2$
$\log_3 (-2-1)+\log_3 (2(-2)+1)\stackrel{?}{=}2$
The logarithms are not defined for $x=-2$.
$x_2=\dfrac{5}{2}$
$\log_3 \left(\dfrac{5}{2}-1\right)+\log_3 \left(2\cdot\dfrac{5}{2}+1\right)\stackrel{?}{=}2$
$\log_3 \dfrac{3}{2}+\log_3 6\stackrel{?}{=}2$
$\log_3 \left(\dfrac{3}{2}\cdot 6\right)\stackrel{?}{=}2$
$\log_3 9\stackrel{?}{=}2$
$2=2\checkmark$
The solution set is:
$\left\{\dfrac{5}{2}\right\}$
