Let's go through each logarithm one by one.

1) $\log_5(5)$

• By the definition of logarithms, $\log_b(b) = 1$ for any base $b \neq 0$.
• Therefore, $\log_5(5) = 1$.

2) $\log_9(729)$

• We need to express 729 as a power of 9. Notice that $9 = 3^2$ and $729 = 3^6$. Therefore, we have: $729 = (3^2)^3 = 9^3$
• Thus, $\log_9(729) = 3$.

3) $\log_{3/4}(64/27)$

• We need to express $64/27$ as a power of $3/4$. Notice that: $64 = 4^3 \quad \text{and} \quad 27 = 3^3$ So, $\frac{64}{27} = \left(\frac{4}{3}\right)^3 = \left(\frac{3}{4}\right)^{-3}$
• Therefore, $\log_{3/4}(64/27) = -3$.

4) $\log_4(2)$

• We want to express 2 as a power of 4. Notice that $2 = 4^{1/2}$.
• Therefore, $\log_4(2) = \frac{1}{2}$.

5) $\log_5(1)$

• By the property of logarithms, $\log_b(1) = 0$ for any base $b \neq 0$.
• Thus, $\log_5(1) = 0$.

Final Answers:

1. $\log_5(5) = 1$
2. $\log_9(729) = 3$
3. $\log_{3/4}(64/27) = -3$
4. $\log_4(2) = \frac{1}{2}$
5. $\log_5(1) = 0$

Would you like more details on any step or have further questions?

Here are 5 related questions to expand your understanding:

1. What is $\log_7(49)$?
2. Simplify $\log_2(16)$.
3. How would you calculate $\log_3(81)$?
4. What is $\log_6(36)$?
5. Can you evaluate $\log_8(64)$?

Tip: Understanding how to rewrite numbers as powers of the base will help in simplifying logarithmic expressions quickly.