Answer
The only solution is $ x=2.5$
Work Step by Step
We can write the expression as a single logarithm using the rule:
$\log_b(m)+\log_b(n)=\log_b(m\cdot n)$
So,
$\log_3(x-1)+\log_3(2x+1)=2$
$\log_3\left((x-1)(2x+1)\right)=2$
$2x^2+x-2x-1=3^2$
$2x^2-x-1=9$
$2x^2-x-10=0$
$(2x-5)(x+2)=0$
$x_1+2=0$
$x_1=-2$
$2x_2-5=0$
$2x_2=5$
$x_2=2.5$
Check for extraneous solutions:
$\log_3(-2-1)+\log_3(2(-2)+1)=2$
$\log_3(-3)+\log_3(-3)=2$
Taking the logarithm of a negative number is not possible; thus x=-2 isn't a solution.
$\log_3(2.5-1)+\log_3(2(2.5)+1)=2$
$\log_3(1.5)+\log_3(6)=2$
$\log_3(1.5\cdot6)=2$
$\log_3(9)=2$
$2=2\checkmark$
