Chapter 6.6, Problem 34E

To determine

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

Answer to Problem 34E

The required real value of $x$ is $9$ .

Explanation of Solution

Given Information:

The given equation is ${\log }_{6}3x+{\log }_6\left(x-1\right)=3$ .

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 ${\log }_{6}3x+{\log }_6\left(x-1\right)=3$ .

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

  ${\log }_{6}3x\left(x-1\right)=3$

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

  $3x\left(x-1\right)=6^3$

Simplify the equation.

  $\begin{array}{c}3x\left(x-1\right)=6^3\\ 3x^2-3x=216\\ 3x^2-3x-216=0\\ 3\left(x^2-x-72\right)=0\end{array}$

Simplify the above equation.

  $\begin{array}{c}3\left(x^2-x-72\right)=0\\ x^2-x-72=0\end{array}$

Now, use quadratic formula in the above equation $x^2-x-72=0$ .

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $1$ for $a$ , $-1$ for $b$ and $-72$ for $c$ in the above equation.

  $x=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(1\right)\left(-72\right)}}{2\left(1\right)}$

Simplify the equation.

  $\begin{array}{c}x=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(1\right)\left(-72\right)}}{2\left(1\right)}\\ =\frac{1\pm \sqrt{1+288}}{2}\\ =\frac{1\pm \sqrt{289}}{2}\\ =\frac{1\pm \sqrt{{17}^2}}{2}\end{array}$

Simplify the equation.

  $\begin{array}{c}x=\frac{1\pm \sqrt{{17}^2}}{2}\\ =\frac{1\pm 17}{2}\\ =\frac{1+17}{2},\frac{1-17}{2}\\ =9,-8\end{array}$

Check for extraneous solution:

For $x=9$

  ${\log }_{6}3x+{\log }_6\left(x-1\right)=3$

Substitute $9$ for $x$ in the given above equation.

  ${\log }_{6}3\left(9\right)+{\log }_6\left(9-1\right)=3$

Simplify the equation.

  $\begin{array}{c}{\log }_{6}3\left(9\right)+{\log }_6\left(9-1\right)=3\\ {\log }_{6}27+{\log }_{6}8=3\end{array}$

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

  ${\log }_6\left(27\right)\left(8\right)=3$

Simplify the equation.

  $\begin{array}{c}{\log }_{6}216=3\\ {\log }_6{6}^3=3\end{array}$

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

  $3{\log }_{6}6=3$

Substitute $1$ for ${\log }_{6}6$ in the above equation.

  $\begin{array}{c}3\left(1\right)=3\\ 3=3\end{array}$

For $x=-8$

  ${\log }_{6}3x+{\log }_6\left(x-1\right)=3$

Substitute $-8$ for $x$ in the given above equation.

  ${\log }_{6}3\left(-8\right)+{\log }_6\left(-8-1\right)=3$

Simplify the equation.

  $\begin{array}{c}{\log }_{6}3\left(-8\right)+{\log }_6\left(-8-1\right)=3\\ {\log }_6\left(-24\right)+{\log }_2\left(-9\right)=3\end{array}$

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

Thus, at $x=-8$ the given function has extraneous solution.

Hence, the required real value of $x$ is $9$ .