Which of the following real numbers when simplified are neither terminating nor repeating decimal ?

Options

A. `sin75^(@)*cos75^(@)`
B. `log_(2)28`
C. `log_(3)5*log_(5)6`
D. `8^(-(log_(27)3))`

Answer

To determine which of the given real numbers, when simplified, are neither terminating nor repeating decimals, we will analyze each option step by step.

### Step 1: Analyze the first option
**Option 1: \( \sin(75^\circ) \cdot \cos(75^\circ) \)**

1. We can use the identity \( 2 \sin A \cos A = \sin(2A) \).
2. Therefore, \( \sin(75^\circ) \cdot \cos(75^\circ) = \frac{1}{2} \sin(150^\circ) \).
3. Now, \( \sin(150^\circ) = \sin(30^\circ) = \frac{1}{2} \).
4. Thus, \( \sin(75^\circ) \cdot \cos(75^\circ) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \).
5. The decimal representation of \( \frac{1}{4} \) is \( 0.25 \), which is a terminating decimal.

**Conclusion for Option 1:** This option is not correct.

### Step 2: Analyze the second option
**Option 2: \( \log_2(28) \)**

1. Let \( x = \log_2(28) \).
2. By the definition of logarithms, this means \( 2^x = 28 \).
3. The value of \( x \) is not an integer because \( 2^4 = 16 \) and \( 2^5 = 32 \), so \( x \) lies between 4 and 5.
4. Since \( x \) is not an integer, \( \log_2(28) \) cannot be expressed as a terminating or repeating decimal.

**Conclusion for Option 2:** This option is correct.

### Step 3: Analyze the third option
**Option 3: \( \log_3(5) \cdot \log_5(6) \)**

1. We can apply the change of base formula: \( \log_a(b) = \frac{\log_c(b)}{\log_c(a)} \).
2. Thus, \( \log_3(5) \) and \( \log_5(6) \) will yield fractional values because they do not simplify to integers.
3. The product of two non-integer logarithms will also be a non-integer, meaning it cannot be a terminating or repeating decimal.

**Conclusion for Option 3:** This option is correct.

### Step 4: Analyze the fourth option
**Option 4: \( 8^{- \log_{27}(3)} \)**

1. We can rewrite \( \log_{27}(3) \) using the change of base formula: \( \log_{27}(3) = \frac{\log(3)}{\log(27)} \).
2. Since \( 27 = 3^3 \), we have \( \log(27) = 3 \log(3) \).
3. Thus, \( \log_{27}(3) = \frac{\log(3)}{3 \log(3)} = \frac{1}{3} \).
4. Therefore, \( 8^{- \log_{27}(3)} = 8^{-\frac{1}{3}} = \frac{1}{8^{\frac{1}{3}}} = \frac{1}{2} \).
5. The decimal representation of \( \frac{1}{2} \) is \( 0.5 \), which is a terminating decimal.

**Conclusion for Option 4:** This option is not correct.

### Final Conclusion
The options that yield neither terminating nor repeating decimals are:
- **Option 2: \( \log_2(28) \)**
- **Option 3: \( \log_3(5) \cdot \log_5(6) \)**