Chapter 6.6, Problem 12E

To determine

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

Answer to Problem 12E

The required value of $x$ is $-\frac{5}{3}$ .

Explanation of Solution

Given Information:

The given equation is ${512}^{5x-1}={\left(\frac{1}{8}\right)}^{-4-x}$ .

Formula Used Power rule of logarithmic functions:

  ${\log }_a{x}^p=p{\log }_{a}x$

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

Calculation:

Consider the equation ${512}^{5x-1}={\left(\frac{1}{8}\right)}^{-4-x}$ .

Simplify the above equation.

  $\begin{array}{c}{512}^{5x-1}={\left(\frac{1}{8}\right)}^{-4-x}\\ {\left\{{\left(8\right)}^3\right\}}^{\left(5x-1\right)}={\left\{{\left(8\right)}^{-1}\right\}}^{\left(-4-x\right)}\\ 8^{\left(15x-3\right)}=8^{\left(x+4\right)}\end{array}$

Take logarithm each side with base $10$ of the above equation $8^{\left(15x-3\right)}=8^{\left(x+4\right)}$ .

  ${\log }_{10}{8}^{\left(15x-3\right)}={\log }_{10}{8}^{\left(x+4\right)}$

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

  $\left(15x-3\right){\log }_{10}8=\left(x+4\right){\log }_{10}8$

Divide each side by ${\log }_{10}8$ in the above equation.

  $\begin{array}{c}\frac{\left(15x-3\right){\log }_{10}8}{{\log }_{10}8}=\frac{\left(x+4\right){\log }_{10}8}{{\log }_{10}8}\\ 15x-3=x+4\\ 15x-x=4+3\\ 14x=7\end{array}$

Simplify the above equation.

  $\begin{array}{c}14x=7\\ x=\frac{7}{14}\\ x=\frac{1}{2}\end{array}$

Hence, the required value of $x$ is $\frac{1}{2}$ .