The number of digits in ` 20^(301)` (given `log_(10) 2 = 0.3010` ) is

Options

A. 602
B. 301
C. 392
D. 391

Answer

To find the number of digits in \( 20^{301} \), we can use the formula for the number of digits \( d \) in a number \( n \), which is given by:

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

In this case, \( n = 20^{301} \). Therefore, we need to calculate \( \log_{10} (20^{301}) \).

### Step 1: Use the logarithmic property
Using the property of logarithms that states \( \log_b (a^c) = c \cdot \log_b a \):

\[ \log_{10} (20^{301}) = 301 \cdot \log_{10} (20) \]

### Step 2: Express \( 20 \) in terms of its prime factors
We can express \( 20 \) as:

\[ 20 = 2 \times 10 = 2 \times (2 \times 5) = 2^2 \times 5 \]

### Step 3: Calculate \( \log_{10} (20) \)
Using the logarithmic property for products, we have:

\[ \log_{10} (20) = \log_{10} (2^2 \times 5) = \log_{10} (2^2) + \log_{10} (5) \]

Using the property \( \log_b (a^c) = c \cdot \log_b a \):

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

### Step 4: Substitute the known value
We know that \( \log_{10} (2) = 0.3010 \):

\[ \log_{10} (2^2) = 2 \cdot 0.3010 = 0.6020 \]

Now, we need to find \( \log_{10} (5) \). We can use the fact that:

\[ \log_{10} (10) = \log_{10} (2 \times 5) = \log_{10} (2) + \log_{10} (5) \]

Since \( \log_{10} (10) = 1 \):

\[ 1 = 0.3010 + \log_{10} (5) \]

Thus,

\[ \log_{10} (5) = 1 - 0.3010 = 0.6990 \]

### Step 5: Calculate \( \log_{10} (20) \)
Now we can calculate \( \log_{10} (20) \):

\[ \log_{10} (20) = 0.6020 + 0.6990 = 1.3010 \]

### Step 6: Substitute back into the logarithm for \( 20^{301} \)
Now substitute \( \log_{10} (20) \) back into our equation for \( \log_{10} (20^{301}) \):

\[ \log_{10} (20^{301}) = 301 \cdot 1.3010 = 391.3010 \]

### Step 7: Calculate the number of digits
Now we can find the number of digits:

\[ d = \lfloor 391.3010 \rfloor + 1 = 391 + 1 = 392 \]

Thus, the number of digits in \( 20^{301} \) is **392**.

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