Chapter 8, Problem 13CR

To determine

To calculate: The solution of the equation ${\log }_3\left(x+8\right)+{\log }_{3}x=2$

Answer to Problem 13CR

Solution:

The solution of the equation ${\log }_3\left(x+8\right)+{\log }_{3}x=2$ is $x=1$

Explanation of Solution

Given information:

The equation ${\log }_3\left(x+8\right)+{\log }_{3}x=2$

Formula used:

Addition rule of logarithm ${\log }_c\left(a\right)+{\log }_c\left(b\right)={\log }_c\left(ab\right)$

Calculation:

Consider the equation ${\log }_3\left(x+8\right)+{\log }_{3}x=2$

By using the addition rule of logarithm

$\Rightarrow {\log }_3\left((x+8)\cdot x\right)=2$

$\Rightarrow {\log }_3\left((x+8)\cdot x\right)=2$

$\Rightarrow {\log }_3\left(x^2+8x\right)=2$

By definition of logarithms, if ${\log }_a\left(b\right)=c$, then $b=a^c$

$\Rightarrow x^2+8x=3^2$

$\Rightarrow x^2+8x=9$.

Subtract $9$ from both sides

$\Rightarrow x^2+8x-9=0$.

Factor out the above equation

$\Rightarrow \left(x-1\right)\left(x+9\right)=0$

$\Rightarrow \left(x-1\right)=0$ or $\left(x+9\right)=0$

$\Rightarrow x=1$ or $x=-9$.

Logarithms that do not have negative solutions

Therefore, $x=1$ is solution of ${\log }_3\left(x+8\right)+{\log }_{3}x=2$