Chapter 6.6, Problem 39E

To determine

To calculate: the value of $x$ and check for extraneous solution.

Answer to Problem 39E

The required real value of $x$ are $3\left(2+\sqrt{2}\right)$ and $3\left(2-\sqrt{2}\right)$ .

Explanation of Solution

Given Information:

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

Formula Used:

Product rule of logarithm:

  ${\log }_{a}x+{\log }_{a}y={\log }_a\left(xy\right)$

Power rule of logarithm:

  ${\log }_a{x}^p=p{\log }_{a}x$

Property of logarithms:

If ${\log }_a\left(x\right)=b$ then $x=a^b$

Quadratic formula:

If $a{x}^2+bx+c=0$ then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Where $a,b,c$ are constant and $a\ne 0$ .

Calculation:

Consider the equation is ${\log }_3\left(x-9\right)+{\log }_3\left(x-3\right)=2$ .

Use product rule of logarithm ${\log }_{a}x+{\log }_{a}y={\log }_a\left(xy\right)$ in the above equation.

  ${\log }_3\left(x-9\right)\left(x-3\right)=2$

Now, use the property of logarithms of the above equation.

  $\begin{array}{c}\left(x-9\right)\left(x-3\right)=3^2\\ x^2-3x-9x+27-9=0\\ x^2-12x+18=0\end{array}$

Now, use quadratic formula in the above equation $x^2-12x+18=0$ .

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $1$ for $a$ , $-12$ for $b$ and $18$ for $c$ in the above equation.

  $x=\frac{-\left(-12\right)\pm \sqrt{{\left(-12\right)}^2-4\left(1\right)\left(18\right)}}{2\left(1\right)}$

Simplify the above equation.

  $\begin{array}{c}x=\frac{-\left(-12\right)\pm \sqrt{{\left(-12\right)}^2-4\left(1\right)\left(18\right)}}{2\left(1\right)}\\ =\frac{12\pm \sqrt{144-72}}{2}\\ =\frac{12\pm \sqrt{72}}{2}\\ =\frac{12\pm \sqrt{6^2\times 2}}{2}\end{array}$

Simplify the equation.

  $\begin{array}{c}x=\frac{12\pm \sqrt{6^2\times 2}}{2}\\ =\frac{12\pm 6\sqrt{2}}{2}\\ =\frac{12+6\sqrt{2}}{2},\frac{12-6\sqrt{2}}{2}\\ =\frac{6\left(2+\sqrt{2}\right)}{2},\frac{6\left(2-\sqrt{2}\right)}{2}\end{array}$

Simplify the equation.

  $\begin{array}{c}x=\frac{6\left(2+\sqrt{2}\right)}{2},\frac{6\left(2-\sqrt{2}\right)}{2}\\ =3\left(2+\sqrt{2}\right),3\left(2-\sqrt{2}\right)\end{array}$

Check for extraneous solution.

For $x=3\left(2+\sqrt{2}\right)$ .

  ${\log }_3\left(x-9\right)+{\log }_3\left(x-3\right)=2$

Substitute $3\left(2+\sqrt{2}\right)$ for $x$ in the given above equation.

  ${\log }_3\left[3\left(2+\sqrt{2}\right)-9\right]+{\log }_3\left[3\left(2+\sqrt{2}\right)-3\right]=2$

Simplify the above equation.

  $\begin{array}{c}{\log }_3\left[3\left(2+\sqrt{2}\right)-9\right]+{\log }_3\left[3\left(2+\sqrt{2}\right)-3\right]=2\\ {\log }_3\left[6+3\sqrt{2}-9\right]+{\log }_3\left[6+3\sqrt{2}-3\right]=2\\ {\log }_3\left[3\sqrt{2}-3\right]+{\log }_3\left[3\sqrt{2}+3\right]=2\end{array}$

Use product rule of logarithm ${\log }_{a}x+{\log }_{a}y={\log }_a\left(xy\right)$ in the above equation.

  ${\log }_3\left[3\sqrt{2}-3\right]\left[3\sqrt{2}+3\right]=2$

Simplify the above equation.

  $\begin{array}{c}{\log }_3\left[3\sqrt{2}-3\right]\left[3\sqrt{2}+3\right]=2\\ {\log }_3\left[9\left(2\right)+9\sqrt{2}-9\sqrt{2}-9\right]=2\\ {\log }_3\left[18-9\right]=2\\ {\log }_{3}9=2\end{array}$

Simplify the equation.

  $\begin{array}{c}{\log }_{3}9=2\\ {\log }_3{3}^2=2\end{array}$

Now, use power rule of logarithm ${\log }_a{x}^p=p{\log }_{a}x$ in the above equation.

  $\begin{array}{c}{\log }_3{3}^2=2\\ 2{\log }_{3}3=2\end{array}$

Substitute $1$ for ${\log }_{3}3$ in the above equation.

  $\begin{array}{c}2\left(1\right)=2\\ 2=2\end{array}$

For $x=3\left(2-\sqrt{2}\right)$ .

  ${\log }_3\left(x-9\right)+{\log }_3\left(x-3\right)=2$

Substitute $3\left(2-\sqrt{2}\right)$ for $x$ in the given above equation.

  ${\log }_3\left[3\left(2-\sqrt{2}\right)-9\right]+{\log }_3\left[3\left(2-\sqrt{2}\right)-3\right]=2$

Simplify the above equation.

  $\begin{array}{c}{\log }_3\left[3\left(2-\sqrt{2}\right)-9\right]+{\log }_3\left[3\left(2-\sqrt{2}\right)-3\right]=2\\ {\log }_3\left[6-3\sqrt{2}-9\right]+{\log }_3\left[6-3\sqrt{2}-3\right]=2\\ {\log }_3\left[-3\sqrt{2}-3\right]+{\log }_3\left[-3\sqrt{2}+3\right]=2\end{array}$

Use product rule of logarithm ${\log }_{a}x+{\log }_{a}y={\log }_a\left(xy\right)$ in the above equation.

  ${\log }_3\left[-3\sqrt{2}-3\right]\left[-3\sqrt{2}+3\right]=2$

Simplify the above equation.

  $\begin{array}{c}{\log }_3\left[-3\sqrt{2}-3\right]\left[-3\sqrt{2}+3\right]=2\\ {\log }_3\left[9\left(2\right)-9\sqrt{2}+9\sqrt{2}-9\right]=2\\ {\log }_3\left[18-9\right]=2\\ {\log }_{3}9=2\end{array}$

Simplify the equation.

  $\begin{array}{c}{\log }_{3}9=2\\ {\log }_3{3}^2=2\end{array}$

Now, use power rule of logarithm ${\log }_a{x}^p=p{\log }_{a}x$ in the above equation.

  $\begin{array}{c}{\log }_3{3}^2=2\\ 2{\log }_{3}3=2\end{array}$

Substitute $1$ for ${\log }_{3}3$ in the above equation.

  $\begin{array}{c}2\left(1\right)=2\\ 2=2\end{array}$

Thus, the given equation has no extraneous solution.

Hence, the required real values of $x$ are $3\left(2+\sqrt{2}\right)$ and $3\left(2-\sqrt{2}\right)$ .