Question:
What is the least integer value of n such that \(\frac{1}{2^n} < 0.001\)?
What is the least integer value of n such that \(\frac{1}{2^n} < 0.001\)?
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It's highly beneficial to memorize powers of 2 up to \(2^{10}\) for standardized tests, as they appear frequently in various contexts, including computer science, logarithms, and inequalities like this one.
Updated On: Oct 1, 2025
- 10
- 11
- 500
- 501
- there is no such least value.
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
This question asks for the smallest integer \(n\) that satisfies a given inequality involving an exponential term.
Step 2: Key Formula or Approach:
The strategy is to rewrite the decimal as a fraction and then solve the inequality.
Step 3: Detailed Explanation:
1. Convert the decimal to a fraction. \[ 0.001 = \frac{1}{1000} \] 2. Rewrite the inequality. The inequality becomes: \[ \frac{1}{2^n} < \frac{1}{1000} \] 3. Solve the inequality. Since the numerators are both 1 and all terms are positive, we can take the reciprocal of both sides. When we do this, the direction of the inequality sign reverses. \[ 2^n > 1000 \] 4. Find the least integer n. We need to find the smallest integer power of 2 that is greater than 1000. It is helpful to know some powers of 2. - \(2^1 = 2\) - \(2^2 = 4\) - ... - \(2^5 = 32\) - ... - \(2^9 = 512\) - \(2^{10} = 1024\) We see that \(2^9 = 512\), which is not greater than 1000. The next integer power, \(2^{10} = 1024\), is the first one that is greater than 1000.
Step 4: Final Answer:
The least integer value of n that satisfies the condition is 10.
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