If log2=0.301 and log3=0.477, find the number of integers in the number of zeroes after the decimal is `3^(-500)` .

Answer

To solve the problem of finding the number of integers in the number of zeros after the decimal in \(3^{-500}\), we can follow these steps:

### Step 1: Define the variable
Let \( a = 3^{-500} \).

### Step 2: Take the logarithm
Taking the logarithm of both sides, we have:
\[ \log a = \log(3^{-500}) \]

### Step 3: Apply the power rule of logarithms
Using the power rule of logarithms, we can simplify this to:
\[ \log a = -500 \log 3 \]

### Step 4: Substitute the value of \(\log 3\)
We know from the problem that \(\log 3 = 0.477\). Therefore, we substitute this value into the equation:
\[ \log a = -500 \times 0.477 \]

### Step 5: Calculate \(-500 \times 0.477\)
Now, we perform the multiplication:
\[ \log a = -238.5 \]

### Step 6: Determine the number of zeros after the decimal
The number of zeros after the decimal point in \(a\) can be found by determining the integer part of \(-\log a\). Since \(\log a = -238.5\), we have:
\[ -\log a = 238.5 \]
The integer part of \(238.5\) is \(238\). Therefore, the number of integers in the number of zeros after the decimal in \(3^{-500}\) is:
\[ \lfloor 238.5 \rfloor = 238 \]

### Final Answer
Thus, the number of integers in the number of zeros after the decimal in \(3^{-500}\) is \(238\).

---