Chapter 6.6, Problem 71E

To determine

To calculate: .The equation for the value of $x$ .

Answer to Problem 71E

The required value of $x$ is $2$ .

Explanation of Solution

Given Information:

The given equation is $2^{2x}-12\cdot 2^x+32=0$ .

Formula Used:

Power rule of logarithmic functions.

  ${\log }_a{x}^p=p{\log }_{a}x$

Where $a$ is the base, $x$ is the variable and $p$ is the power of variable.

Property of logarithm.

  ${\log }_{a}a=1$

Calculation:

Consider the equation is $2^{2x}-12\cdot 2^2+32=0$ .

Simplify the above equation.

  $\begin{array}{c}{2}^{2x}-12\cdot 2^2+32=0\\ 2^{2x}-48+32=0\\ 2^{2x}-16=0\\ 2^{2x}=16\end{array}$

Now, $2^{2x}=16$

Take logarithm each side with base $2$ .

  $\begin{array}{l}{\log }_2{2}^{2x}={\log }_{2}16\\ {\log }_2{2}^{2x}={\log }_2{2}^4\end{array}$

Now, use power rule of logarithm ${\log }_a{x}^p=p{\log }_{a}x$ in the above equation.

  $2x{\log }_{2}2=4{\log }_{2}2$

Use property of logarithm ${\log }_{a}a=1$ in the above equation.

  $\begin{array}{c}2x\left(1\right)=4\left(1\right)\\ 2x=4\end{array}$

Simplify the above equation.

  $\begin{array}{c}2x=4\\ x=\frac{4}{2}\\ x=2\end{array}$

Hence, the required value of $x$ is $2$ .