Given log10 2 = 0.30103 and log10 3 = 0.47712, find the number of digits in 3^12 × 2^8.

Difficulty: Easy

Correct Answer: 9

Explanation:

Introduction / Context:We apply the digit-count formula using given logarithms for 2 and 3. The product's log is the sum of individual logs, scaled by exponents.

Given Data / Assumptions:

  • log10 2 = 0.30103, log10 3 = 0.47712.
  • N = 3^12 × 2^8.

Concept / Approach:

  • log10 N = 12 log10 3 + 8 log10 2.
  • Digits = floor(log10 N) + 1.

Step-by-Step Solution:

log10 N = 12(0.47712) + 8(0.30103) = 5.72544 + 2.40824 = 8.13368Digits = floor(8.13368) + 1 = 8 + 1 = 9

Verification / Alternative check:Since 10^8 < N < 10^9, there must be 9 digits.

Why Other Options Are Wrong:

  • 6, 7, 8 underestimate the magnitude.

Final Answer:9

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