Question

Solve:${\log }_2(3x-2)+{\log }_{2}x=4$

Short answer

The solution of the given equation:$x=-1,\frac{8}{3}$

Step-by-step solution

Step 1. Given information

Given equation:

${\log }_2(3x-2)+{\log }_{2}x=4$

Step 2. Solve the equation.

$$\begin{gathered} {\log }_2(3x-2)+{\log }_{2}x=4 \\ {\log }_2(3x-2)\left(x\right)=4 \end{gathered}$$

Take antilog on both sides.

$$\begin{gathered} x(3x-2)=2^4 \\ 3x^2-2x=16 \\ 3x^2-2x-16=0 \\ 3x^2-8x+6x-16=0 \\ x(3x-8)+2(3x-8)=0 \\ (3x-8)(x+2)=0 \\ 3x-8=0 \\ x=\frac{8}{3} \\ x+2=0 \\ x=-2 \end{gathered}$$