Let's solve each part step by step:

a. $\log_2 (\sqrt[4]{2})$

We need to evaluate $\log_2 (\sqrt[4]{2})$, which means the logarithm base 2 of the fourth root of 2.

We know that: $\sqrt[4]{2} = 2^{1/4}$

Thus, the expression becomes: $\log_2 (2^{1/4})$

Using the logarithmic property $\log_b (a^n) = n \log_b a$, we get: $\log_2 (2^{1/4}) = \frac{1}{4} \log_2 (2)$

Since $\log_2 (2) = 1$, we have: $\frac{1}{4} \times 1 = \frac{1}{4}$

So, the value of $\log_2 (\sqrt[4]{2})$ is $\frac{1}{4}$.

b. $\log_{125} (25)$

We need to evaluate $\log_{125} (25)$, which means the logarithm base 125 of 25.

First, express 125 and 25 as powers of 5: $125 = 5^3 \quad \text{and} \quad 25 = 5^2$

Thus, the expression becomes: $\log_{5^3} (5^2)$

Using the change of base formula: $\log_{a^n} (a^m) = \frac{m}{n}$

We get: $\log_{5^3} (5^2) = \frac{2}{3}$

So, the value of $\log_{125} (25)$ is $\frac{2}{3}$.

Final Answers:

• a. $\frac{1}{4}$
• b. $\frac{2}{3}$

Would you like more details on these calculations or have any questions?

Related Questions:

1. What are the properties of logarithms?
2. How do you solve logarithmic equations?
3. What is the change of base formula in logarithms?
4. How can logarithmic functions be graphed?
5. How do fractional exponents work with logarithms?

Tip: When solving logarithms with roots, remember to rewrite roots as fractional exponents for easier simplification!