If k is an integer and 10^(k)-1 , then (k =

1. Translate the problem requirements: We need to find integer k such that \(10^{(k-1)} < 0.000125 < 10^k\). This means 0.000125 falls between two consecutive powers of 10.
2. Convert decimal to scientific notation: Express 0.000125 in the form \(a \times 10^n\) to identify which power of 10 it's closest to.
3. Identify the power of 10 range: Determine which consecutive powers of 10 bracket our decimal value.
4. Apply the inequality constraints: Use the identified range to solve for k using the given inequality \(10^{(k-1)} < 0.000125 < 10^k\).

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this inequality is asking us. We have: \(10^{(k-1)} < 0.000125 < 10^k\)

This means we need to find an integer k where 0.000125 sits between two consecutive powers of 10. Think of it like this: we're looking for which "power of 10 neighborhood" our decimal lives in.

The left side, \(10^{(k-1)}\), represents the smaller power of 10, and the right side, \(10^k\), represents the larger power of 10. Since these are consecutive integer powers, 0.000125 must fall somewhere between them.

Process Skill: TRANSLATE - Converting the inequality language into a clear range-finding problem

2. Convert decimal to scientific notation

To understand where 0.000125 fits, let's convert it to scientific notation by counting decimal places.

Starting with 0.000125, let's move the decimal point to get a number between 1 and 10:

- \(0.000125 = 0.00125 \times 10^{-1}\)
- \(0.000125 = 0.0125 \times 10^{-2}\)
- \(0.000125 = 0.125 \times 10^{-3}\)
- \(0.000125 = 1.25 \times 10^{-4}\)

So \(0.000125 = 1.25 \times 10^{-4}\)

This tells us that 0.000125 is 1.25 times \(10^{-4}\). Since 1.25 is between 1 and 10, this confirms our scientific notation is correct.

3. Identify the power of 10 range

Now we need to figure out which consecutive powers of 10 bracket our value.

Since \(0.000125 = 1.25 \times 10^{-4}\), and 1.25 is greater than 1, we know:

- 0.000125 is greater than \(1 \times 10^{-4} = 10^{-4}\)
- 0.000125 is less than \(10 \times 10^{-4} = 10^{-3}\)

Let's verify this makes sense:

- \(10^{-4} = 0.0001\)
- \(10^{-3} = 0.001\)

Indeed: \(0.0001 < 0.000125 < 0.001\) ✓

So our decimal falls between \(10^{-4}\) and \(10^{-3}\).

4. Apply the inequality constraints

Now we match our range to the given inequality format: \(10^{(k-1)} < 0.000125 < 10^k\)

We found that: \(10^{-4} < 0.000125 < 10^{-3}\)

Comparing the exponents:

- The smaller power: \(k-1 = -4\)
- The larger power: \(k = -3\)

From \(k = -3\), we can verify: \(k-1 = -3-1 = -4\) ✓

Let's double-check our work:

- \(10^{(k-1)} = 10^{-4} = 0.0001\)
- \(10^k = 10^{-3} = 0.001\)
- Indeed: \(0.0001 < 0.000125 < 0.001\) ✓

Process Skill: APPLY CONSTRAINTS - Using the inequality structure to solve for k

5. Final Answer

Therefore, \(k = -3\).

Looking at our answer choices, this corresponds to choice D: -3.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the inequality structure: Students may not recognize that \(10^{(k-1)} < 0.000125 < 10^k\) represents finding consecutive powers of 10 that bracket the decimal. They might try to solve each inequality separately instead of understanding this as a range-finding problem where the decimal falls between two specific powers of 10.

2. Confusion about which power is smaller: Students may incorrectly assume that since k-1 appears on the left, it represents the larger value, when in fact \(10^{(k-1)}\) must be the smaller bound since it has the smaller exponent. This leads to setting up the problem backwards.

Errors while executing the approach

1. Scientific notation conversion errors: When converting 0.000125 to scientific notation, students often miscount decimal places or move the decimal in the wrong direction. A common error is writing \(0.000125 = 1.25 \times 10^{-3}\) instead of the correct \(1.25 \times 10^{-4}\), leading to an incorrect final answer.

2. Incorrect power of 10 calculations: Students may struggle with negative exponents, confusing \(10^{-3} = 0.001\) with \(10^{-4} = 0.0001\). They might also incorrectly calculate that 0.000125 falls between \(10^{-3}\) and \(10^{-2}\) instead of the correct range between \(10^{-4}\) and \(10^{-3}\).

3. Arithmetic errors in decimal comparisons: When verifying that \(0.0001 < 0.000125 < 0.001\), students may make basic comparison errors, especially when dealing with multiple zeros after the decimal point, leading them to incorrect conclusions about which powers of 10 bracket the given decimal.

Errors while selecting the answer

1. Confusing k and k-1 in final substitution: After correctly identifying that \(10^{-4} < 0.000125 < 10^{-3}\), students may incorrectly assign \(k = -4\) instead of \(k = -3\), forgetting that k should equal the larger exponent (-3) and k-1 should equal the smaller exponent (-4).