Chapter 6.6, Problem 37E

To determine

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

Answer to Problem 37E

The required real value of $x$ is $\pm 1$ .

Explanation of Solution

Given Information:

The given equation is ${\log }_{3}3x^2+{\log }_{3}3=2$ .

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 $x=a^b$ .

Calculation:

Consider the equation ${\log }_{3}3x^2+{\log }_{3}3=2$ .

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

  ${\log }_3\left(3x^2\right)\left(3\right)=2$

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

  $\begin{array}{c}3x^2\left(3\right)=3^2\\ 9x^2=9\end{array}$

Divide each side by $9$ in the above equation $9x^2=9$ .

  $\begin{array}{c}\frac{9x^2}{9}=\frac{9}{9}\\ x^2=1\\ x=\sqrt{1}\\ x=\pm 1\end{array}$

Check for extraneous solution.

For $x=1$ .

  ${\log }_{3}3x^2+{\log }_{3}3=2$

Substitute $1$ for $x$ in the given above equation.

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

Substitute $1$ for ${\log }_{3}3$ in the above equation.

  $\begin{array}{c}1+1=2\\ 2=2\end{array}$

For $x=-1$

  ${\log }_{3}3x^2+{\log }_{3}3=2$

Substitute $-1$ for $x$ in the given above equation.

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

Substitute $1$ for ${\log }_{3}3$ in the above equation.

  $\begin{array}{c}1+1=2\\ 2=2\end{array}$

Thus, the given equation has no extraneous.

Hence, the required value of $x$ is $\pm 1$ .