Answer
$\left\{\dfrac{8}{3}\right\}$
Work Step by Step
We are given the equation:
$\log_2 (3x-2)+\log_2 x=4$
Use the Product Property of logarithms:
$\log_2 x(3x-2)=4$
Rewrite the equation in exponential form:
$x(3x-2)=2^4$
$3x^2-2x-16=0$
Solve the equation:
$x=\dfrac{2\pm\sqrt{(-2)^2-4(3)(-16)}}{2(3)}=\dfrac{2\pm 14}{6}=\dfrac{1\pm 7}{3}$
$x_1=\dfrac{1-7}{3}=-2$
$x_2=\dfrac{1+7}{3}=\dfrac{8}{3}$
Check the solutions:
$x=-2$
$\log_2 (3(-2)-2)+\log_2 (-2)\stackrel{?}{=}4$
The logarithms are undefined for negative values!
$x=\dfrac{8}{3}$
$\log_2 \left(3\left(\dfrac{8}{3}\right)-2\right)+\log_2 \dfrac{8}{3}\stackrel{?}{=}4$
$\log_2 6+\log_2 \dfrac{8}{3}\stackrel{?}{=}4$
$\log_2 \left(6\cdot\dfrac{8}{3}\right)\stackrel{?}{=}4$
$\log_2 16\stackrel{?}{=}4$
$4=4\checkmark$
The solution set is:
$\left\{\dfrac{8}{3}\right\}$
