Chapter 6.6, Problem 36E

To determine

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

Answer to Problem 36E

The required real value of $x$ are $13.2$ .

Explanation of Solution

Given Information:

The given equation is $\ln x+\ln \left(x-2\right)=5$ .

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 $\ln x+\ln \left(x-2\right)=5$ .

It is known that the base ofnatural logarithm is $e$ .

Rewrite the given equation with base $e$ .

  ${\ln }_{e}x+{\ln }_e\left(x-2\right)=5$

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

  ${\ln }_{e}x\left(x-2\right)=5$

Now, use the property of logarithms of the above equation ${\ln }_{e}x\left(x-2\right)=5$ .

  $\begin{array}{c}x\left(x-2\right)=e^5\\ x^2-2x-e^5=0\end{array}$

Now, use quadratic formula in the above equation $x^2-2x-e^5=0$ .

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $1$ for $a$ , $-2$ for $b$ and $-e^5$ for $c$ in the above equation.

  $x=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(1\right)\left(-e^5\right)}}{2\left(1\right)}$

Simplify the equation.

  $\begin{array}{c}x=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(1\right)\left(-e^5\right)}}{2\left(1\right)}\\ =\frac{2\pm \sqrt{4+4e^4}}{2}\end{array}$

It is known that the value of $e$ is $2.718$ .

Substitute $2.718$ for $e$ in the above equation $x=\frac{2\pm \sqrt{4+4e^4}}{2}$ .

  $x=\frac{2\pm \sqrt{4+4e^5}}{2}$

Simplify the above equation.

  $\begin{array}{c}x=\frac{2\pm \sqrt{4+4e^5}}{2}\\ =\frac{2\pm \sqrt{4+4{\left(2.718\right)}^5}}{2}\\ =\frac{2\pm \sqrt{4+4\left(148.336\right)}}{2}\\ =\frac{2\pm \sqrt{4+593.344}}{2}\end{array}$

Simplify the equation.

  $\begin{array}{c}x=\frac{2\pm \sqrt{4+593.344}}{2}\\ =\frac{2\pm \sqrt{597.344}}{2}\\ =\frac{2\pm 24.4}{2}\\ =\frac{2+24.4}{2},\frac{2-24.4}{2}\end{array}$

Simplify the above equation.

  $\begin{array}{c}x=\frac{2+24.4}{2},\frac{2-24.4}{2}\\ =\frac{26.4}{2},\frac{-22.4}{2}\\ =13.2,-11.2\end{array}$

Check for extraneous solution.

For $x=13.2$ .

  $\ln x+\ln \left(x-2\right)=5$

Substitute $13.2$ for $x$ in the given above equation.

  $\begin{array}{c}\ln \left(13.2\right)+\ln \left(13.2-2\right)=5\\ \ln \left(13.2\right)+\ln \left(11.2\right)=5\end{array}$

Substitute $-11.2$ for $\ln \left(13.2\right)$ and $2.4$ for $\ln \left(11.2\right)$ in the above equation.

  $\begin{array}{c}2.6+2.4=5\\ 5=5\end{array}$

For $x=-11.2$ .

  $\ln x+\ln \left(x-2\right)=5$

Substitute $-11.2$ for $x$ in the given above equation.

  $\begin{array}{c}\ln \left(-11.2\right)+\ln \left(-11.2-2\right)=5\\ \ln \left(-11.2\right)+\ln \left(-13.2\right)=5\end{array}$

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

Thus, at $x=-11.2$ the given equation has extraneous solution.

Hence, the required real values of $x$ are $13.2$ .