Chapter 11, Problem 5CR

To determine

The solution of the equation ${\log }_3\left(x-1\right)+{\log }_3\left(2x+1\right)=2$.

Answer to Problem 5CR

Solution:

The solution of the equation ${\log }_3\left(x-1\right)+{\log }_3\left(2x+1\right)=2$ is $x=\frac{5}{2}$.

Explanation of Solution

Given information:

The equation ${\log }_3\left(x-1\right)+{\log }_3\left(2x+1\right)=2$.

Explanation:

Consider the equation, ${\log }_3\left(x-1\right)+{\log }_3\left(2x+1\right)=2$.

The domain of the variable requires that $x-1>0$ and $2x+1>0$, so $x>1$ and $x>-\frac{1}{2}$, that is equivalently $x>-\frac{1}{2}$.

This means, the solution of the equation must satisfy $x>-\frac{1}{2}$.

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

The equation ${\log }_3\left(x-1\right)+{\log }_3\left(2x+1\right)=2$ becomes,

${\log }_3\left(x-1\right)\left(2x+1\right)=2$

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

$\Rightarrow \left(x-1\right)\left(2x+1\right)=3^2$

$\Rightarrow 2x^2+x-2x-1=9$

$\Rightarrow 2x^2-x-10=0$

By using quadratic formula,

$x=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(2\right)\left(-10\right)}}{2\left(2\right)}$

$\Rightarrow x=\frac{1\pm \sqrt{81}}{4}$

$\Rightarrow x=\frac{1\pm 9}{4}$

$\Rightarrow x=\frac{1+9}{4}\text{or}x=\frac{1-9}{4}$

$\Rightarrow x=\frac{10}{4}\text{or}x=\frac{-8}{4}$

$\Rightarrow x=\frac{5}{2}\text{or}x=-2$

Here $x=-2$ do not satisfy $x>-\frac{1}{2}$.

Thus $x=\frac{5}{2}$ is the solution of the equation ${\log }_3\left(x-1\right)+{\log }_3\left(2x+1\right)=2$.