Chapter 5, Problem 45RE

To determine

To Find: The single logarithm form of each of the expression.

Answer to Problem 45RE

The single logarithm form of the given equation is $\frac{25}{4}{\log }_{4}x$ .

Explanation of Solution

Given:

The given expression is $3{\log }_4{x}^2+\frac{1}{2}{\log }_4\sqrt{x}$ .

Calculation:

Consider the given expression is,

  $3{\log }_4{x}^2+\frac{1}{2}{\log }_4\sqrt{x}$

Consider the property of logarithm is,

  $c{\log }_{a}b={\log }_a{b}^c$

Then,

  $\begin{array}{c}3{\log }_4{x}^2+\frac{1}{2}{\log }_4\sqrt{x}={\log }_4{x}^{2\left(3\right)}+{\log }_4{\sqrt{x}}^{\frac{1}{2}}\\ ={\log }_4{x}^6+{\log }_4{x}^{\frac{1}{4}}\end{array}$

Consider the logarithm property,

  ${\log }_{b}a+{\log }_{b}c={\log }_{b}ac$

Then,

  $\begin{array}{c}3{\log }_4{x}^2+\frac{1}{2}{\log }_4\sqrt{x}={\log }_4{x}^6+{\log }_4{x}^{\frac{1}{4}}\\ ={\log }_4\left(x^6+x^{\frac{1}{4}}\right)\\ ={\log }_4\left(x^{6+\frac{1}{4}}\right)\\ ={\log }_4{x}^{\frac{25}{4}}\end{array}$

Solve further as,

  $3{\log }_4{x}^2+\frac{1}{2}{\log }_4\sqrt{x}=\frac{25}{4}{\log }_{4}x$