Chapter 6.6, Problem 40E

To determine

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

Answer to Problem 40E

The required real value of $x$ are $2.72$ .

Explanation of Solution

Given Information:

The given equation is ${\log }_5\left(x+4\right)+{\log }_5\left(x+1\right)=2$ .

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 ${\log }_5\left(x+4\right)+{\log }_5\left(x+1\right)=2$ .

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

  ${\log }_5\left(x+4\right)\left(x+1\right)=2$

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

  $\left(x+4\right)\left(x+1\right)=5^2$

Simplify the above equation.

  $\begin{array}{c}\left(x+4\right)\left(x+1\right)=5^2\\ x^2+x+4x+4=25\\ x^2+5x+4-25=0\\ x^2+5x-21=0\end{array}$

Now, use quadratic formula in the above equation $x^2+5x-21=0$ .

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $1$ for $a$ , $5$ for $b$ and $-21$ for $c$ in the above equation.

  $x=\frac{-5\pm \sqrt{5^2-4\left(1\right)\left(-21\right)}}{2\left(1\right)}$

Simplify the above equation.

  $\begin{array}{c}x=\frac{-5\pm \sqrt{5^2-4\left(1\right)\left(-21\right)}}{2\left(1\right)}\\ =\frac{-5\pm \sqrt{25+84}}{2}\\ =\frac{-5\pm \sqrt{109}}{2}\\ =\frac{-5\pm 10.44}{2}\end{array}$

Simplify the equation.

  $\begin{array}{l}x=\frac{-5\pm 10.44}{2}\\ =\frac{-5+10.44}{2},\frac{-5-10.44}{2}\\ =\frac{5.44}{2},\frac{-15.44}{2}\\ =2.72,-7.72\end{array}$

Check for extraneous solution.

For $x=2.72$ .

  ${\log }_5\left(x+4\right)+{\log }_5\left(x+1\right)=2$

Substitute $2.72$ for $x$ in the given above equation.

  ${\log }_5\left(x+4\right)+{\log }_5\left(x+1\right)=2$

Simplify the above equation.

  $\begin{array}{c}{\log }_5\left(x+4\right)+{\log }_5\left(x+1\right)=2\\ {\log }_5\left(2.72+4\right)+{\log }_5\left(2.72+1\right)=2\\ {\log }_5\left(6.72\right)+{\log }_5\left(6.72\right)=2\end{array}$

Substitute $1.18$ for ${\log }_5\left(6.72\right)$ and $0.82$ for ${\log }_5\left(6.72\right)$ in the above equation.

  $\begin{array}{c}1.18+0.82=2\\ 2=2\end{array}$

For $x=-7.72$ .

  ${\log }_5\left(x+4\right)+{\log }_5\left(x+1\right)=2$

Substitute $-7.72$ for $x$ in the given above equation.

  ${\log }_5\left(x+4\right)+{\log }_5\left(x+1\right)=2$

Simplify the above equation.

  $\begin{array}{c}{\log }_5\left(x+4\right)+{\log }_5\left(x+1\right)=2\\ {\log }_5\left(-7.72+4\right)+{\log }_5\left(-7.72+1\right)=2\\ {\log }_5\left(-3.72\right)+{\log }_5\left(-6.72\right)=2\end{array}$

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

Thus, at $x=-7.72$ the given equation has extraneous solution.

Hence, the required real values of $x$ are $2.72$ .