Chapter 5, Problem 14CR

To determine

To calculate: The solution of the equation ${\log }_3\left(x+1\right)+{\log }_3\left(2x-3\right)={\log }_9\left(9\right)$ .

Answer to Problem 14CR

Solution:

The solution of the equation ${\log }_3\left(x+1\right)+{\log }_3\left(2x-3\right)={\log }_9\left(9\right)$ is $\underset{\_}{x=2}$ .

Explanation of Solution

Given information:

The equation ${\log }_3\left(x+1\right)+{\log }_3\left(2x-3\right)={\log }_9\left(9\right)$ .

Formula used:

1) ${\log }_c\left(c\right)=1$ .

2) Sum rule of logarithm: ${\log }_c\left(a\right)+{\log }_c\left(b\right)={\log }_c\left(ab\right)$ .

3) ${\log }_a\left(b\right)=c\iff b=a^c$ .

Calculation:

As ${\log }_c\left(c\right)=1$ , then ${\log }_9\left(9\right)=1$ ,

  $\Rightarrow {\log }_3\left(x+1\right)+{\log }_3\left(2x-3\right)=1$

By using sum rule of logarithm,

  $\Rightarrow {\log }_3\left(\left(x+1\right)\left(2x-3\right)\right)=1$

Using distributive property,

  $\Rightarrow {\log }_3\left(2x^2-3x+2x-3\right)=1$

Simplify,

  $\Rightarrow {\log }_3\left(2x^2-x-3\right)=1$

Changing logarithm statement to exponential,

  $\Rightarrow 2x^2-x-3=3^1$

  $\Rightarrow 2x^2-x-3=3$

Subtract $3$ from both sides,

  $\Rightarrow 2x^2-x-6=0$

For the quadratic equation $a{x}^2+bx+c=0$ , solutions are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Comparing equation $2x^2-x-6=0$ with $a{x}^2+bx+c=0$ gives $a=2,b=-1,c=-6$ .

Substitute the values $a=2,b=-1,c=-6$ in the quadratic formula.

The solutions of equation $2x^2-x-6=0$ are $x=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(2\right)\left(-6\right)}}{2\left(2\right)}$ ,

  $\Rightarrow x=\frac{1\pm \sqrt{1+48}}{4}$

  $\Rightarrow x=\frac{1\pm \sqrt{49}}{4}$

  $\Rightarrow x=\frac{1\pm 7}{4}$

  $\Rightarrow x=\frac{1+7}{4},x=\frac{1-7}{4}$

  $\Rightarrow x=\frac{8}{4},x=\frac{-6}{4}$

  $\Rightarrow x=2,x=-\frac{3}{2}$

Verify the solutions for equation ${\log }_3\left(x+1\right)+{\log }_3\left(2x-3\right)=1$ .

Substitute $x=2$ in an equation ${\log }_3\left(x+1\right)+{\log }_3\left(2x-3\right)=1$ ,

  $\Rightarrow {\log }_3\left(2+1\right)+{\log }_3\left(2\left(2\right)-3\right)=1$

  $\Rightarrow {\log }_3\left(3\right)+{\log }_3\left(4-3\right)=1$

  $\Rightarrow 1+{\log }_3\left(1\right)=1$

  $\Rightarrow 1+0=1$

  $\Rightarrow 1=1$

So, left side and right sides are equal. Therefore, $x=2$ is the solution of the equation ${\log }_3\left(x+1\right)+{\log }_3\left(2x-3\right)={\log }_9\left(9\right)$ .

Now, substitute $x=-\frac{3}{2}$ in the equation ${\log }_3\left(x+1\right)+{\log }_3\left(2x-3\right)=1$ ,

  $\Rightarrow {\log }_3\left(-\frac{3}{2}+1\right)+{\log }_3\left(2\left(-\frac{3}{2}\right)-3\right)=1$

  $\Rightarrow {\log }_3\left(-\frac{1}{2}\right)+{\log }_3\left(-3-3\right)=1$

Log of negative numbers is undefined. So, ${\log }_3\left(-\frac{1}{2}\right)$ is undefined. Therefore, $x=-\frac{3}{2}$ is not the solution of equation ${\log }_3\left(x+1\right)+{\log }_3\left(2x-3\right)={\log }_9\left(9\right)$ .

Therefore, the solution of equation ${\log }_3\left(x+1\right)+{\log }_3\left(2x-3\right)={\log }_9\left(9\right)$ is $x=2$ .