Chapter 3, Problem 14CT

To determine

To expand the expression

  ${\log }_2\frac{5\sqrt{x}}{6}$

Answer to Problem 14CT

  ${\log }_2\frac{5\sqrt{x}}{6}={\log }_{2}5+\frac{1}{2}{\log }_{2}x-{\log }_{2}6$

Explanation of Solution

Given:

  ${\log }_2\frac{5\sqrt{x}}{6}$

Concept Used:

For $a\in R^{+}\&a\ne 1$ and $u\in R^{+}$ :

Power property of logarithmic function: ${\log }_a{u}^n=n{\log }_{a}u$

Product property of logarithmic function: ${\log }_{a}uv={\log }_{a}u+{\log }_{a}v$

Quotient property of logarithmic function: ${\log }_a\frac{u}{v}={\log }_{a}u-{\log }_{a}v$

Calculation:

To expand the expression

  ${\log }_2\frac{5\sqrt{x}}{6}$

Simplify this using the properties of logarithmic function ${\log }_{a}uv={\log }_{a}u+{\log }_{a}v$ and ${\log }_a\frac{u}{v}={\log }_{a}u-{\log }_{a}v$ .

As,

  $\begin{array}{c}{\log }_2\frac{5\sqrt{x}}{6}={\log }_{2}5\sqrt{x}-{\log }_{2}6\\ ={\log }_{2}5+{\log }_2\sqrt{x}-{\log }_{2}6\\ ={\log }_{2}5+{\log }_2{x}^{1/2}-{\log }_{2}6\end{array}$

Now to further expand use the property ${\log }_a{u}^n=n{\log }_{a}u$ .

As,

  $\begin{array}{c}{\log }_2\frac{5\sqrt{x}}{6}={\log }_{2}5+{\log }_2{x}^{1/2}-{\log }_{2}6\\ ={\log }_{2}5+\frac{1}{2}{\log }_{2}x-{\log }_{2}6\end{array}$

Thus,

  ${\log }_2\frac{5\sqrt{x}}{6}={\log }_{2}5+\frac{1}{2}{\log }_{2}x-{\log }_{2}6$