Chapter 6.6, Problem 33E

To determine

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

Answer to Problem 33E

The required real value of $x$ is $4$ .

Explanation of Solution

Given Information:

The given equation is ${\log }_{2}x+{\log }_2\left(x-2\right)=3$ .

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 ${\log }_{2}x+{\log }_2\left(x-2\right)=3$ .

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

  ${\log }_{2}x\left(x-2\right)=3$

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

  $x\left(x-2\right)=2^3$

Simplify the equation.

  $\begin{array}{c}x\left(x-2\right)=2^3\\ x^2-2x=8\\ x^2-2x-8=0\end{array}$

Now, use quadratic formula in the above equation $x^2-2x-8=0$ .

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $1$ for $a$ , $-2$ for $b$ and $-8$ for $c$ in the above equation.

  $x=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(1\right)\left(-8\right)}}{2\left(1\right)}$

Simplify the equation.

  $\begin{array}{c}x=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(1\right)\left(-8\right)}}{2\left(1\right)}\\ =\frac{2\pm \sqrt{4+32}}{2}\\ =\frac{2\pm \sqrt{36}}{2}\\ =\frac{2\pm \sqrt{6^2}}{2}\end{array}$

Simplify the above equation.

  $\begin{array}{c}x=\frac{2\pm \sqrt{6^2}}{2}\\ =\frac{2\pm 6}{2}\\ =\frac{2+6}{2},\frac{2-6}{2}\\ =4,-2\end{array}$

Check for extraneous solution:

For $x=4$

  ${\log }_{2}x+{\log }_2\left(x-2\right)=3$

Substitute $4$ for $x$ in the given above equation.

  ${\log }_{2}4+{\log }_2\left(4-2\right)=3$

Simplify the equation.

  $\begin{array}{c}{\log }_{2}4+{\log }_2\left(4-2\right)=3\\ {\log }_2{\left(2\right)}^2+{\log }_{2}2=3\end{array}$

Use power rule of logarithm in the above equation ${\log }_2{\left(2\right)}^2+{\log }_{2}2=3$ .

  $2{\log }_{2}2+{\log }_{2}2=3$ .

Substitute ${\log }_{2}2$ for ${\log }_{2}2$ in the above equation.

  $\begin{array}{l}2{\log }_{2}2+{\log }_{2}2=3\\ 2\left(1\right)+\left(1\right)=3\\ 2+1=3\\ 3=3\end{array}$

For $x=-2$

  ${\log }_{2}x+{\log }_2\left(x-2\right)=3$

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

  ${\log }_2\left(-2\right)+{\log }_2\left(-2-2\right)=3$

Simplify the equation.

  $\begin{array}{c}{\log }_2\left(-2\right)+{\log }_2\left(-2-2\right)=3\\ {\log }_2\left(-2\right)+{\log }_2\left(-4\right)=3\end{array}$

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

Thus, at $x=-2$ the given equation has extraneous solution.

Hence, the required real value of $x$ is $4$ .