Chapter 8.2, Problem 32PPS

To determine

To calculate: The value of x in the equation ${\left(\frac{1}{2}\right)}^{4x+1}=8^{2x+1}$ .

Answer to Problem 32PPS

The value of x in the equation ${\left(\frac{1}{2}\right)}^{4x+1}=8^{2x+1}$ is $-\frac{2}{5}$ .

Explanation of Solution

Given information:

The given equation is ${\left(\frac{1}{2}\right)}^{4x+1}=8^{2x+1}$ .

Formula used:

Property of equality of exponential functions:

If $b^x=b^y$ then $x=y$for all $b>0$ ,and $b\ne 1$ .

Calculation:

Consider the given equation,

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

Since 8 can be written as $2^3$ and $\frac{1}{2}$ can be written as $2^{-1}$ .So, rewrite the above equation as like this,

  $\begin{array}{c}{\left(\frac{1}{2}\right)}^{4x+1}=8^{2x+1}\\ 2^{-\left(4x+1\right)}=2^{3\left(2x+1\right)}\\ 2^{-4x-1}=2^{6x+3}\end{array}$

Now, apply the property of equality of exponential functions:

  $\begin{array}{c}-4x-1=6x+3\\ -4x-6x=3+1\\ -10x=4\\ x=-\frac{4}{10}\end{array}$

Proceed to next,

  $\begin{array}{l}x=-\frac{4}{10}\\ x=-\frac{2}{5}\end{array}$

Therefore, the value of x in the equation ${\left(\frac{1}{2}\right)}^{4x+1}=8^{2x+1}$ is $-\frac{2}{5}$ .