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

Answer

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

### Step 1: Define the variable
Let \( y = 6^{20} \).

### Step 2: Take the logarithm
We take the logarithm (base 10) of both sides:
\[ \log_{10} y = \log_{10} (6^{20}) \]

### Step 3: Use the power rule of logarithms
Using the property of logarithms that states \( \log_{10} (m^n) = n \cdot \log_{10} m \), we can rewrite the equation:
\[ \log_{10} y = 20 \cdot \log_{10} 6 \]

### Step 4: Express \( 6 \) in terms of its prime factors
We know that \( 6 = 2 \times 3 \). Therefore, we can express \( \log_{10} 6 \) as:
\[ \log_{10} 6 = \log_{10} (2 \times 3) = \log_{10} 2 + \log_{10} 3 \]

### Step 5: Substitute the values of \( \log_{10} 2 \) and \( \log_{10} 3 \)
Given that \( \log_{10} 2 = 0.301 \) and \( \log_{10} 3 = 0.477 \), we substitute these values:
\[ \log_{10} 6 = 0.301 + 0.477 = 0.778 \]

### Step 6: Substitute back into the equation
Now we substitute \( \log_{10} 6 \) back into our equation for \( \log_{10} y \):
\[ \log_{10} y = 20 \cdot 0.778 = 15.56 \]

### Step 7: Convert logarithm back to the original number
Using the property that if \( \log_{10} m = x \), then \( m = 10^x \), we find:
\[ y = 10^{15.56} \]

### Step 8: Determine the number of integers
To find the number of integers in \( y \), we take the floor of \( 15.56 \) and add 1:
\[ \text{Number of integers} = \lfloor 15.56 \rfloor + 1 = 15 + 1 = 16 \]

Thus, the final answer is:
\[ \text{The number of integers in } 6^{20} \text{ is } 16. \]

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