If log 2 (to base 10) is 0.30103, how many digits are there in the number 2^56?

Introduction / Context:
This question checks your understanding of how to use logarithms to determine the number of digits in a large power of a number. Instead of directly computing 2^56, which is unwieldy, we use the relationship between common logarithms and decimal digit counts.

Given Data / Assumptions:

• log 2 = 0.30103 (base 10).
• We need the number of digits in 2^56.
• All logs are common logarithms (base 10).

Concept / Approach:
If a positive integer N has common logarithm log N = L, then the number of digits of N in base 10 is given by floor(L) + 1. We therefore compute log(2^56) using laws of logarithms, then apply this digit formula. The key rules are log(a^n) = n log a and the floor plus one rule for digit count.

Step-by-Step Solution:
We want the number of digits in N = 2^56. Compute log N = log(2^56) = 56 log 2. Given log 2 = 0.30103, we have log N = 56 × 0.30103. Multiply: 0.30103 × 50 = 15.0515 and 0.30103 × 6 = 1.80618. Add them: 15.0515 + 1.80618 ≈ 16.85768. So log(2^56) ≈ 16.85768. The number of digits in N is floor(16.85768) + 1 = 16 + 1 = 17.

Verification / Alternative check:
We know 2^10 ≈ 10^3 (since 2^10 = 1024). Then 2^50 ≈ 10^15 and 2^6 = 64. So 2^56 = 2^50 × 2^6 ≈ 10^15 × 64 ≈ 6.4 × 10^16, which clearly has 17 digits. This quick estimation matches the detailed logarithmic computation.

Why Other Options Are Wrong:
19, 23, 25: These values would correspond to much larger exponents. For example, a 19-digit number would be around 10^18, while our estimate shows 2^56 is of the order 10^16 to 10^17. Therefore all these options are too large for the digit count.

Common Pitfalls:
Some learners mistakenly take the integer part of the logarithm without adding one, which would give 16 instead of 17. Others miscalculate 56 × 0.30103 or round too roughly. It is important to remember the precise rule: number of digits = floor(log₁₀ N) + 1 for positive integers.

Final Answer:
The number 2^56 has 17 digits.