Chapter 5, Problem 20CT

To determine

${\log }_2\left(\frac{4x^3}{x^2-3x-18}\right)$ as the sum and/or difference of logarithms. Express powers as factors.

Answer to Problem 20CT

Solution:

The simplified expression of ${\log }_2\left(\frac{4x^3}{x^2-3x-18}\right)$ is $2+3{\log }_2\left(x\right)-{\log }_2\left(x-6\right)-{\log }_2\left(x+3\right)$.

Explanation of Solution

Given information:

The expressionis ${\log }_2\left(\frac{4x^3}{x^2-3x-18}\right)$.

Explanation:

Consider the expression ${\log }_2\left(\frac{4x^3}{x^2-3x-18}\right)$.

By using ${\log }_{a}M-{\log }_{a}N={\log }_a\left(\frac{M}{N}\right)$,

$\Rightarrow {\log }_2\left(\frac{4x^3}{x^2-3x-18}\right)={\log }_2\left(4x^3\right)-{\log }_2\left(x^2-3x-18\right)$

$\Rightarrow {\log }_2\left(\frac{4x^3}{x^2-3x-18}\right)={\log }_2\left(4\cdot x^3\right)-{\log }_2\left(x-6\right)\left(x+3\right)$

By using ${\log }_{a}M+{\log }_{a}N={\log }_a\left(MN\right)$,

$\Rightarrow {\log }_2\left(\frac{4x^3}{x^2-3x-18}\right)={\log }_2\left(4\right)+{\log }_2\left(x^3\right)-\left[{\log }_2\left(x-6\right)+{\log }_2\left(x+3\right)\right]$

$\Rightarrow {\log }_2\left(\frac{4x^3}{x^2-3x-18}\right)={\log }_2\left(2^2\right)+{\log }_2\left(x^3\right)-\left[{\log }_2\left(x-6\right)+{\log }_2\left(x+3\right)\right]$

By using ${\log }_a{M}^N=N{\log }_{a}M$,

$\Rightarrow {\log }_2\left(\frac{4x^3}{x^2-3x-18}\right)={\log }_2\left(2^2\right)+{\log }_2\left(x^3\right)-\left[{\log }_2\left(x-6\right)+{\log }_2\left(x+3\right)\right]$

$\Rightarrow {\log }_2\left(\frac{4x^3}{x^2-3x-18}\right)=2{\log }_2\left(2\right)+3{\log }_2\left(x\right)-{\log }_2\left(x-6\right)-{\log }_2\left(x+3\right)$

By using ${\log }_{a}a=1$,

$\Rightarrow {\log }_2\left(\frac{4x^3}{x^2-3x-18}\right)=2\cdot 1+3{\log }_2\left(x\right)-{\log }_2\left(x-6\right)-{\log }_2\left(x+3\right)$

$\Rightarrow {\log }_2\left(\frac{4x^3}{x^2-3x-18}\right)=2+3{\log }_2\left(x\right)-{\log }_2\left(x-6\right)-{\log }_2\left(x+3\right)$

Therefore, the simplified expression of ${\log }_2\left(\frac{4x^3}{x^2-3x-18}\right)$ is $2+3{\log }_2\left(x\right)-{\log }_2\left(x-6\right)-{\log }_2\left(x+3\right)$.