Chapter 6.6, Problem 38E

To determine

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

Answer to Problem 38E

The required real value of $x$ are $-2$ and $-8$ .

Explanation of Solution

Given Information:

The given equation is ${\log }_4\left(-x\right)+{\log }_4\left(x+10\right)=2$ .

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 ${\log }_4\left(-x\right)+{\log }_4\left(x+10\right)=2$ .

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

  ${\log }_4\left(-x\right)\left(x+10\right)=2$

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

  $\left(-x\right)\left(x+10\right)=4^2$

Simplify the above equation.

  $\begin{array}{c}\left(-x\right)\left(x+10\right)=4^2\\ -x^2-10x=16\\ -x^2-10x-16=0\end{array}$

Now, use quadratic formula in the above equation $-x^2-10x-16=0$ .

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $-1$ for $a$ , $-10$ for $b$ and $-16$ for $c$ in the above equation.

  $x=\frac{-\left(-10\right)\pm \sqrt{{\left(-10\right)}^2-4\left(-1\right)\left(-16\right)}}{2\left(-1\right)}$

Simplify the above equation.

  $\begin{array}{c}x=\frac{-\left(-10\right)\pm \sqrt{{\left(-10\right)}^2-4\left(-1\right)\left(-16\right)}}{2\left(-1\right)}\\ =\frac{10\pm \sqrt{100-64}}{-2}\\ =\frac{10\pm \sqrt{36}}{-2}\\ =\frac{10\pm \sqrt{6^2}}{-2}\end{array}$

Simplify the above equation.

  $\begin{array}{c}x=\frac{10\pm \sqrt{6^2}}{-2}\\ =\frac{10\pm 6}{-2}\\ =\frac{10+6}{-2},\frac{10-6}{-2}\\ =-8,-2\end{array}$

Check for extraneous solution.

For $x=-2$ .

  ${\log }_4\left(-x\right)+{\log }_4\left(x+10\right)=2$

Substitute $-2$ for $x$ in the given above equation.

  $\begin{array}{c}{\log }_4\left\{-\left(-2\right)\right\}+{\log }_4\left(-2+10\right)=2\\ {\log }_{4}2+{\log }_{4}8=2\end{array}$

Use product rule of logarithm in the above equation.

  ${\log }_4\left(2\right)\left(8\right)=2$

Simplify the above equation.

  $\begin{array}{c}{\log }_4\left(2\right)\left(8\right)=2\\ {\log }_{4}16=2\\ {\log }_4{4}^2=2\end{array}$

Now, use power rule of logarithm in the above equation.

  $\begin{array}{c}{\log }_4{4}^2=2\\ 2{\log }_{4}4=2\end{array}$

Substitute $1$ for ${\log }_{4}4$ in the above equation.

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

For $x=-8$ .

  ${\log }_4\left(-x\right)+{\log }_4\left(x+10\right)=2$

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

  $\begin{array}{c}{\log }_4\left\{-\left(-8\right)\right\}+{\log }_4\left(-8+10\right)=2\\ {\log }_4\left(8\right)+{\log }_4\left(2\right)=2\end{array}$

Use product rule of logarithm in the above equation.

  ${\log }_4\left(8\right)\left(2\right)=2$

Simplify the equation.

  $\begin{array}{c}{\log }_4\left(8\right)\left(2\right)=2\\ {\log }_{4}16=2\\ {\log }_4{4}^2=2\end{array}$

Now, use power rule of logarithm in the above equation.

  $\begin{array}{c}{\log }_4{4}^2=2\\ 2{\log }_{4}4=2\end{array}$

Substitute $1$ for ${\log }_{4}4$ 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 $-2$ and $-8$ .