Chapter 6.6, Problem 35E

To determine

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

Answer to Problem 35E

The required real value of $x$ are $6.04$ .

Explanation of Solution

Given Information:

The given equation is $\ln x+\ln \left(x+3\right)=4$ .

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 $\ln x+\ln \left(x+3\right)=4$ .

It is known that the base of natural logarithm is $e$ .

Rewrite the equation with base $e$ .

  ${\ln }_{e}x+{\ln }_e\left(x+3\right)=4$

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

  ${\ln }_{e}x\left(x+3\right)=4$

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

  $\begin{array}{c}x\left(x+3\right)=e^4\\ x^2+3x-e^4=0\end{array}$

Now, use quadratic formula in the above equation $x^2+3x-e^4=0$ .

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $1$ for $a$ , $3$ for $b$ and $-e^4$ for $c$ in the above equation.

  $x=\frac{-3\pm \sqrt{3^2-4\left(1\right)\left(-e^4\right)}}{2\left(1\right)}$

Simplify the equation.

  $\begin{array}{c}x=\frac{-3\pm \sqrt{3^2-4\left(1\right)\left(-e^4\right)}}{2\left(1\right)}\\ =\frac{-3\pm \sqrt{9+4e^4}}{2}\end{array}$

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

Substitute $2.718$ for $e$ in the above equation $x=\frac{-3\pm \sqrt{9+4e^4}}{2}$ .

  $\begin{array}{c}x=\frac{-3\pm \sqrt{9+4e^4}}{2}\\ =\frac{-3\pm \sqrt{9+4{\left(2.718\right)}^4}}{2}\\ =\frac{-3\pm \sqrt{9+4\left(54.576\right)}}{2}\\ =\frac{-3\pm \sqrt{9+218.304}}{2}\end{array}$

Simplify the above equation.

  $\begin{array}{c}x=\frac{-3\pm \sqrt{9+218.304}}{2}\\ =\frac{-3\pm \sqrt{227.304}}{2}\\ =\frac{-3\pm 15.08}{2}\\ =\frac{-3+15.08}{2},\frac{-3-15.08}{2}\end{array}$

Simplify the above equation.

  $\begin{array}{c}x=\frac{-3+15.08}{2},\frac{-3-15.08}{2}\\ =\frac{12.08}{2},\frac{-18.08}{2}\\ =6.04,-9.04\end{array}$

Check for extraneous solution:

For $x=6.04$

  $\ln x+\ln \left(x+3\right)=4$

Substitute $6.04$ for $x$ in the given above equation

  $\begin{array}{c}\ln \left(6.04\right)+\ln \left(6.04+3\right)=4\\ \ln \left(6.04\right)+\ln \left(9.04\right)=4\end{array}$

Substitute $1.8$ for $\ln \left(6.04\right)$ and $2.2$ for $\ln \left(9.04\right)$ in the above equation.

  $\begin{array}{c}1.8+2.2=4\\ 4=4\end{array}$

For $x=-9.04$

  $\ln x+\ln \left(x+3\right)=4$

Substitute $-9.04$ for $x$ in the given above equation

  $\begin{array}{c}\ln \left(-9.04\right)+\ln \left(-9.04+3\right)=4\\ \ln \left(-9.04\right)+\ln \left(-6.04+3\right)=4\end{array}$

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

Thus, at $x=-9.04$ the given equation has extraneous solution.

Hence, the required real values of $x$ are $6.04$ .