Given ,log2=0.301 and log3=0.477, then the number of digits before decimal in `3^12times2^8`is

Options

A. 7
B. 8
C. 9
D. 11

Answer

To find the number of digits before the decimal in the expression \(3^{12} \times 2^{8}\), we can follow these steps:

### Step 1: Define the expression
Let \( a = 3^{12} \times 2^{8} \).

### Step 2: Take the logarithm
We take the logarithm of both sides:
\[ \log a = \log(3^{12} \times 2^{8}). \]

### Step 3: Use logarithm properties
Using the property of logarithms that states \(\log(m \times n) = \log m + \log n\), we can rewrite the equation:
\[ \log a = \log(3^{12}) + \log(2^{8}). \]

### Step 4: Apply the power rule of logarithms
Using the property \(\log(m^n) = n \log m\), we have:
\[ \log a = 12 \log 3 + 8 \log 2. \]

### Step 5: Substitute the given values
We know from the problem statement that \(\log 2 = 0.301\) and \(\log 3 = 0.477\). Substituting these values in gives:
\[ \log a = 12 \times 0.477 + 8 \times 0.301. \]

### Step 6: Calculate each term
Calculating \(12 \times 0.477\):
\[ 12 \times 0.477 = 5.724. \]

Calculating \(8 \times 0.301\):
\[ 8 \times 0.301 = 2.408. \]

### Step 7: Add the results
Now we add the two results together:
\[ \log a = 5.724 + 2.408 = 8.132. \]

### Step 8: Find the number of digits
The number of digits \(d\) in a number \(N\) can be found using the formula:
\[ d = \lfloor \log N \rfloor + 1. \]
Thus, for our case:
\[ d = \lfloor \log a \rfloor + 1 = \lfloor 8.132 \rfloor + 1 = 8 + 1 = 9. \]

### Final Answer
The number of digits before the decimal in \(3^{12} \times 2^{8}\) is **9**.

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