Find the value of the following logarithms.

1) log_5(5) =

2) log_9(729) =

3) log_(3/4)(64/27) =

4) log_4(2) =

5) log_5(1) =

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.

Find the value of the following logarithms.
1) log_5(5) =
2) log_9(729) =
3) log_(3/4)(64/27) =
4) log_4(2) =
5) log_5(1) =

Learn how to evaluate logarithmic expressions such as log_5(5), log_9(729), log_(3/4)(64/27), log_4(2), and log_5(1). This solution breaks down each step with key logarithmic properties and exponent rules. Suitable for students in Grades 9-12.

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