Question

Evaluating Logarithms Evaluate the expression.

(a)${\log }_4\sqrt{2}$ (b) ${\log }_4\left(\frac{1}{2}\right)$ (c)${\log }_{4}8$

Short answer

1. The required value is $\frac{1}{4}$.
2. The required value is$-\frac{1}{2}$.
3. The required value is $\frac{3}{2}$.

Step-by-step solution

Part a. Step 1. Given.

The given expression is ${\log }_4\sqrt{2}$.

Part a. Step 2. To determine.

We have to evaluate the given expression.

Part a. Step 3. Calculation.

$$\begin{gathered} {\log }_4\sqrt{2} \\ ={\log }_4\left(2^{\frac{1}{2}}\right) \\ ={\log }_{2^2}\left(2^{\frac{1}{2}}\right) \\ =\frac{\left(\frac{1}{2}\right)}{2}{\log }_2\left(2\right)\text{[because\hspace{0.17em}\hspace{0.17em}}{\log }_{a^n}\left(b^m\right)=\frac{m}{n}{\log }_a\left(b\right)] \\ =\frac{1}{4}\left(1\right)[\text{because\hspace{0.17em}\hspace{0.17em}}{\log }_a\left(a\right)=1] \\ =\frac{1}{4} \end{gathered}$$

Hence, ${\log }_4\sqrt{2}=\frac{1}{4}$.

Part b. Step 1. Given.

The given exponential form is ${\log }_4\left(\frac{1}{2}\right)$.

Part b. Step 2. To determine.

We have to evaluate the given expression.

Part b. Step 3. Calculation.

$$\begin{gathered} {\log }_4\left(\frac{1}{2}\right) \\ ={\log }_4\left(2^{-1}\right) \\ ={\log }_{2^2}\left(2^{-1}\right) \\ =\frac{\left(-1\right)}{2}{\log }_2\left(2\right)\text{[because\hspace{0.17em}\hspace{0.17em}}{\log }_{a^n}\left(b^m\right)=\frac{m}{n}{\log }_a\left(b\right)] \\ =-\frac{1}{2}\left(1\right)[\text{because\hspace{0.17em}\hspace{0.17em}}{\log }_a\left(a\right)=1] \\ =-\frac{1}{2} \end{gathered}$$

Hence, ${\log }_4\left(\frac{1}{2}\right)=-\frac{1}{2}$.

Part c. Step 1. Given.

The given exponential form is ${\log }_{4}8$.

Part c. Step 2. To determine.

We have to evaluate the given expression.

Part c. Step 3. Calculation.

$$\begin{gathered} {\log }_{4}8 \\ ={\log }_4\left(2^3\right) \\ ={\log }_{2^2}\left(2^3\right) \\ =\frac{3}{2}{\log }_2\left(2\right)\text{[because\hspace{0.17em}\hspace{0.17em}}{\log }_{a^n}\left(b^m\right)=\frac{m}{n}{\log }_a\left(b\right)] \\ =\frac{3}{2}\left(1\right)[\text{because\hspace{0.17em}\hspace{0.17em}}{\log }_a\left(a\right)=1] \\ =\frac{3}{2} \end{gathered}$$

Hence, ${\log }_{4}8=\frac{3}{2}$.