Chapter 6.6, Problem 69E

To determine

To calculate: .The equation for the value of $x$ .

Answer to Problem 69E

The required real value of $x$ is $10.61$

Explanation of Solution

Given Information:

The given equation is ${\log }_3\left(x-6\right)={\log }_{9}2x$ .

Formula Used:

Power rule of logarithm:

  ${\log }_a{x}^p=p{\log }_{a}x$

Cancellation rule of logarithm:

If ${\log }_{a}x={\log }_{a}y$ then $x=y$

Property of logarithms:

  ${\log }_{a}x=\frac{\log x}{\log a}$

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 ${\log }_3\left(x-6\right)={\log }_{9}2x$ .

Now, use the property of logarithms ${\log }_{a}x=\frac{\log x}{\log a}$ of the above equation.

  $\begin{array}{l}{\log }_3\left(x-6\right)=\frac{\log 2x}{\log 9}\\ {\log }_3\left(x-6\right)=\frac{\log 2x}{\log 3^2}\end{array}$

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

  $\begin{array}{l}{\log }_3\left(x-6\right)=\frac{\log 2x}{\log 3^2}\\ {\log }_3\left(x-6\right)=\frac{\log 2x}{2\log 3}\end{array}$

Simplify the equation.

  $\begin{array}{l}{\log }_3\left(x-6\right)=\frac{\log 2x}{2\log 3}\\ 2{\log }_3\left(x-6\right)=\frac{\log 2x}{\log 3}\end{array}$

Use the property of logarithms ${\log }_{a}x=\frac{\log x}{\log a}$ of the above equation.

  $2{\log }_3\left(x-6\right)={\log }_{3}2x$

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

  ${\log }_3{\left(x-6\right)}^2={\log }_{3}2x$

Use cancellation rule of logarithm.

  ${\left(x-6\right)}^2=2x$

Simplify the equation.

  $\begin{array}{l}{\left(x-6\right)}^2=2x\\ x^2-12x+36-2x=0\\ x^2-14x+36=0\end{array}$

Now, use quadratic formula in the above equation $x^2-14x+36=0$ .

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $1$ for $a$ , $-14$ for $b$ and $36$ for $c$ in the above equation.

  $x=\frac{-\left(-14\right)\pm \sqrt{{\left(-14\right)}^2-4\left(1\right)\left(36\right)}}{2\left(1\right)}$

Simplify the above equation.

  $\begin{array}{c}x=\frac{-\left(-14\right)\pm \sqrt{{\left(-14\right)}^2-4\left(1\right)\left(36\right)}}{2\left(1\right)}\\ =\frac{14\pm \sqrt{196-144}}{2}\\ =\frac{14\pm \sqrt{52}}{2}\\ =\frac{14\pm 2\sqrt{13}}{2}\end{array}$

Simplify the equation.

  $\begin{array}{c}x=\frac{14\pm 2\sqrt{13}}{2}\\ =\frac{2\left(7\pm \sqrt{13}\right)}{2}\\ =\left(7\pm \sqrt{13}\right)\end{array}$

Simplify the equation.

  $\begin{array}{c}x=\left(7+\sqrt{13}\right),\left(7-\sqrt{3}\right)\\ =\left(7+3.605\right),\left(7-3.605\right)\\ =10.605,3.395\\ \approx 10.61,3.39\end{array}$

Check for extraneous solution:

For $x=10.61$ .

  ${\log }_3\left(x-6\right)={\log }_{9}2x$

Substitute $10.61$ for $x$ in the given above equation.

  ${\log }_3\left(10.61-6\right)={\log }_{9}2\left(10.61\right)$

Simplify the equation.

  $\begin{array}{c}{\log }_3\left(10.61-6\right)={\log }_{9}2\left(10.61\right)\\ {\log }_3\left(4.61\right)={\log }_9\left(21.22\right)\\ 1.39=1.39\end{array}$

For $x=3.39$ .

  ${\log }_3\left(x-6\right)={\log }_{9}2x$

Substitute $3.39$ for $x$ in the given above equation.

  $\begin{array}{c}{\log }_3\left(3.39-6\right)={\log }_{9}2\left(3.39\right)\\ {\log }_3\left(-2.61\right)={\log }_9\left(6.78\right)\end{array}$

It is known that ${\log }_a\left(-x\right)$ is not exist.

Thus, at $x=3.39$ the given function has extraneous solution.

Hence, the required real value of $x$ is $10.61$ .