How many number of digits are there is `2^(98)` ? Given that `log_(10) 2 = 0.30103`

Options

A. A)98
B. B)99
C. C)30
D. D)29

Answer

To find the number of digits in \(2^{98}\), we can use the formula that relates the number of digits \(d\) of a number \(n\) to its logarithm:

\[ d = \lfloor \log_{10} n \rfloor + 1 \]

In this case, we want to find the number of digits in \(n = 2^{98}\).

### Step 1: Calculate \(\log_{10} (2^{98})\)

Using the logarithmic property that states \(\log_{10} (a^b) = b \cdot \log_{10} a\), we can write:

\[ \log_{10} (2^{98}) = 98 \cdot \log_{10} 2 \]

### Step 2: Substitute the value of \(\log_{10} 2\)

We are given that \(\log_{10} 2 = 0.30103\). Substituting this value into our equation gives:

\[ \log_{10} (2^{98}) = 98 \cdot 0.30103 \]

### Step 3: Perform the multiplication

Now we calculate:

\[ 98 \cdot 0.30103 = 29.50 \]

### Step 4: Find the number of digits

Now, we can find the number of digits using the formula:

\[ d = \lfloor 29.50 \rfloor + 1 \]

Calculating the floor function:

\[ \lfloor 29.50 \rfloor = 29 \]

Thus, the number of digits is:

\[ d = 29 + 1 = 30 \]

### Final Answer

The number of digits in \(2^{98}\) is **30**.

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