Chapter 6.6, Problem 70E

To determine

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

Answer to Problem 70E

The required value of $x$ is $16$ .

Explanation of Solution

Given Information:

The given equation is ${\log }_{4}x={\log }_{8}4x$ .

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.

Cancellation rule of logarithm:

If ${\log }_{a}x={\log }_{a}y$ then $x=y$ .

Property of logarithm:

  ${\log }_{a}x=\frac{\log x}{\log a}$

Calculation:

Consider the equation is ${\log }_{4}x={\log }_{8}4x$ .

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

  $\begin{array}{l}\frac{\log x}{\log 4}=\frac{\log 4x}{\log 8}\\ \frac{\log x}{\log 2^2}=\frac{\log 4x}{\log 2^3}\end{array}$

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

  $\frac{\log x}{2\log 2}=\frac{\log 4x}{3\log 2}$

Simplify the above equation.

  $\begin{array}{l}\frac{\log x}{2\log 2}=\frac{\log 4x}{3\log 2}\\ \frac{3\log x}{\log 2}=\frac{2\log 4x}{\log 2}\end{array}$

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

  $\frac{\log x^3}{\log 2}=\frac{\log {\left(4x\right)}^2}{\log 2}$

Use property of logarithm in above equation.

  ${\log }_2{x}^3={\log }_2{\left(4x\right)}^{{}^2}$

Use cancellation rule of logarithm in the above equation.

  $x^3={\left(4x\right)}^2$

Simplify the equation.

  $\begin{array}{c}{x}^3={\left(4x\right)}^2\\ x^3=16x^2\\ \frac{x^3}{x^2}=16\\ x=16\end{array}$

Hence, the required value of $x$ is $16$ .