If log2=0.301 and log3=0.477, find the number of integers in
(ii) `6^(20)`

Answer

To find the number of integers in \( 6^{20} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
We are given:
\[ \log 2 = 0.301 \quad \text{and} \quad \log 3 = 0.477 \]

2. **Express \( 6^{20} \) in terms of logarithms:**
We can express \( 6^{20} \) using logarithms. Let:
\[ y = 6^{20} \]
Taking logarithm on both sides:
\[ \log y = \log (6^{20}) = 20 \log 6 \]

3. **Break down \( \log 6 \):**
We can express \( 6 \) as \( 2 \times 3 \):
\[ \log 6 = \log (2 \times 3) = \log 2 + \log 3 \]
Substituting the values we have:
\[ \log 6 = \log 2 + \log 3 = 0.301 + 0.477 \]

4. **Calculate \( \log 6 \):**
Now, calculate:
\[ \log 6 = 0.301 + 0.477 = 0.778 \]

5. **Substitute back into the equation for \( \log y \):**
Now substitute \( \log 6 \) back into the equation for \( \log y \):
\[ \log y = 20 \log 6 = 20 \times 0.778 \]

6. **Calculate \( \log y \):**
Now calculate:
\[ \log y = 20 \times 0.778 = 15.56 \]

7. **Convert back to the number \( y \):**
To find \( y \), we can express it in exponential form:
\[ y = 10^{15.56} \]

8. **Determine the number of integers:**
The number of integers in \( y \) can be determined by the integer part of the exponent:
- The integer part of \( 15.56 \) is \( 15 \).
- The next integer greater than \( 15.56 \) is \( 16 \).

Thus, the number of integers in \( 6^{20} \) is \( 16 \).

### Final Answer:
The number of integers in \( 6^{20} \) is \( 16 \).