N is a positive integer with n digits ⟺ n−1 ≤ log₁₀N < n
Example ①
When log₁₀2=0.3010, how many digits does 2^20 have?
Answer: log₁₀(2^20)=20×0.3010=6.020. 6≤6.020<7 → 7 digits
When log₁₀2=0.3010, how many digits does 2^20 have?
Answer: log₁₀(2^20)=20×0.3010=6.020. 6≤6.020<7 → 7 digits
Example ②
When log₁₀2=0.3010 and log₁₀3=0.4771, find the number of digits in 6^10.
Answer: log₁₀(6^10)=10(0.3010+0.4771)=7.781. 7≤7.781<8 → 8 digits
When log₁₀2=0.3010 and log₁₀3=0.4771, find the number of digits in 6^10.
Answer: log₁₀(6^10)=10(0.3010+0.4771)=7.781. 7≤7.781<8 → 8 digits
Key Point
- n digits ⟺ n−1≤log₁₀N<n
- nth decimal place: −n≤log₁₀N<−n+1
- log₁₀2≒0.3010、log₁₀3≒0.4771
Practice Problems
- When log₁₀2=0.3010, find the number of digits in 2^10.
- When log₁₀3=0.4771, at which decimal place does the first nonzero digit of (1/3)^20 appear?
- When log₁₀2=0.3010 and log₁₀7=0.8451, find log₁₀14.
Answers and Explanations
- Answer: 4 digits（log₁₀(2^10)=3.010、3≤3.010<4）
- Answer: 10th decimal place
Explanation: log₁₀(3^(−20))=−9.542. −10≤−9.542<−9. - Answer: 1.1461
Explanation: log₁₀14=log₁₀(2×7)=0.3010+0.8451=1.1461.