Chapter 13, Problem 8CR

To determine

The solution of the equation ${\log }_2\left(3x-2\right)+{\log }_{2}x=4$.

Answer to Problem 8CR

Solution:

The solution of the equation ${\log }_2\left(3x-2\right)+{\log }_{2}x=4$ is $x=\frac{8}{3}$.

Explanation of Solution

Given information:

The equation ${\log }_2\left(3x-2\right)+{\log }_{2}x=4$.

Explanation:

Consider the equation ${\log }_2\left(3x-2\right)+{\log }_{2}x=4$.

The domain of the variable requires that $3x-2>0$ and $x>0$, so $x>\frac{2}{3}$ and $x>0$.

This means, the solution of the equation must satisfy $x>\frac{2}{3}$.

By using addition rule ${\log }_{M}A+{\log }_{M}B={\log }_M\left(AB\right)$,

The equation ${\log }_2\left(3x-2\right)+{\log }_{2}x=4$ becomes

${\log }_2\left(3x-2\right)\left(x\right)=4$

Now, by using if $y={\log }_{a}x$ then $x=a^y,$ where $a>0$ and $a\ne 1,$ are equivalent,

$\Rightarrow \left(3x-2\right)\left(x\right)=2^4$

$\Rightarrow 3x^2-2x=16$

$\Rightarrow 3x^2-2x-16=0$

$\Rightarrow 3x^2-8x+6x-16=0$

$\Rightarrow x\left(3x-8\right)+2(3x-8)=0$

$\Rightarrow \left(3x-8\right)\left(x+2\right)=0$

$\Rightarrow \left(3x-8\right)=0$ or $\left(x+2\right)=0$

$\Rightarrow x=\frac{8}{3}$ or $x=-2$

Here, $x=-2$ does not satisfy $x>\frac{2}{3}$.

Thus, $x=\frac{8}{3}$ is the solution of the equation ${\log }_2\left(3x-2\right)+{\log }_{2}x=4$.