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Bunuel
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 13 Aug 2023, 09:00
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If k is an integer and \(10^{k-1} < 0.000125 < 10^k\), then k =

A. -6
B. -5
C. -4
D. -3
E. -2­


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gmatophobia
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 13 Aug 2023, 09:24
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Bunuel
If k is an integer and \(10^{k-1} < 0.000125 < 10^k\), then k =

A. -6
B. -5
C. -4
D. -2
E. -1

\(10^{k-1} < 0.000125 < 10^k\)

\(10^{k-1} < 125*10^{-6} < 10^k\)

Dividing all sides by \(10^{-6}\)

\(10^{k-1+6} < 125 < 10^{k+6}\)

\(10^{k+5} < 125 < 10^{k+6}\)

Therefore \(k + 5 = 2\)

\(k = -3\)
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 13 Aug 2023, 09:27
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If k is an integer and \(10^{k-1} < 0.000125 < 10^k\), then k =

A. -6
B. -5
C. -4
D. -3
E. -2

We can quickly answer this question by using the answer choices.

Since \(0.000125 < 10^-3\) we can eliminate (E).

Since \(0.000125 > 10^-4\) we can eliminate (A), (B), and (C).

So, the correct answer is (D).
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 13 Aug 2023, 15:04
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Bunuel
If k is an integer and \(10^{k-1} < 0.000125 < 10^k\), then k =

A. -6
B. -5
C. -4
D. -3
E. -2

0.0001 < 0.000125 < 0.001

10ˉ⁴ < 0.000125 < 10ˉ³

k = -3

Answer: D
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 03 Sep 2023, 04:39
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gmatophobia
Bunuel
If k is an integer and \(10^{k-1} < 0.000125 < 10^k\), then k =

A. -6
B. -5
C. -4
D. -2
E. -1

\(10^{k-1} < 0.000125 < 10^k\)

\(10^{k-1} < 125*10^{-6} < 10^k\)

Dividing all sides by \(10^{-6}\)

\(10^{k-1+6} < 125 < 10^{k+6}\)

\(10^{k+5} < 125 < 10^{k+6}\)

Therefore \(k + 5 = 2\)

\(k = -3\)


How did you deduce k + 5 = 2? What am I missing?
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 03 Sep 2023, 04:54
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10^(k−1)<0.000125<10^k
=>10^(k−1)<125*10^(-6)<10^k
=>10^(k−1)<125*10^(-6)<10^k
=>10^(k−1+6)<125<10^(k +6)
=>10^(k+5)<125<10^(k +6)

Since,
=> 100<125<1000
=>10^2<125<10^3

Therefore k+5=2
=>k=-3

Hence D
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 03 Sep 2023, 06:27
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Vineela.mk
How did you deduce k + 5 = 2? What am I missing?

Vineela.mk -

Quote:
\(10^{k+5} < 125 < 10^{k+6}\)

Let's take one part at a time -

\(10^{k+5} < 125 \)

We know k is an integer and that \(10^{k+5} < 125 \). The value, \(10^{k+5}\) is a power of 10. The closest power of 10 which is less than 125 is 100. Therefore we can write the expression as

\(10^{k+5} \leq 100 \)

\(10^{k+5} \leq 10^2 \)

Comparing the powers

\(k + 5 \leq 2\)

\(k \leq -3\)

\(125 < 10^{k+6} \)

The closest power of 10 which is greater than 125 is 1000. Therefore we can write the expression as

\(1000 \leq 10^{k+6} \)

\(10^3 \leq 10^{k+6} \)

Comparing the powers

\(3 \leq k+6 \)

\(-3 \leq k \)

The common value of \(k\) from both ⇒ \(k = -3\)

Hope this helps.
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 06 Sep 2023, 03:12
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Bunuel
If k is an integer and \(10^{k-1} < 0.000125 < 10^k\), then k =

A. -6
B. -5
C. -4
D. -2
E. -1

\(10^{k-1} < 0.000125 < 10^k\)

\(10^{k-1} < 125*10^{-6} < 10^k\)

Dividing all sides by \(10^{-6}\)

\(10^{k-1+6} < 125 < 10^{k+6}\)

\(10^{k+5} < 125 < 10^{k+6}\)

Therefore \(k + 5 = 2\)

\(k = -3\)

After arriving at \(10^{k+5} < 125 < 10^{k+6}\) how did you conclude that \(k + 5 = 2\) ??
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 12 Sep 2023, 11:15
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Hoozan
After arriving at \(10^{k+5} < 125 < 10^{k+6}\) how did you conclude that \(k + 5 = 2\) ??

Hi Hoozan

Checking if you got a chance to refer to this post -

https://gmatclub.com/forum/if-k-is-an-i ... l#p3262194

Let me know if you still have any question.
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = Updated on: 28 Jan 2025, 03:11
Originally posted by KarishmaB on 08 Apr 2024, 09:14.
Last edited by KarishmaB on 28 Jan 2025, 03:11, edited 1 time in total.
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Bunuel
If k is an integer and \(10^{k-1} < 0.000125 < 10^k\), then k =

A. -6
B. -5
C. -4
D. -3
E. -2
­\(10^{k-1} < 0.000125 < 10^k\)

First notice that all options are negative so let's take powers of 10 to the denominator.

\(10^{k-1} < \frac{125}{10^6} < 10^k\)
\(10^{k-1} < \frac{1.25*10^2}{10^6} < 10^k\)
\(10^{k-1} < \frac{ 1.25}{10^4} < 10^k\)

We know that
\(\frac{1}{10^4} < \frac{ 1.25}{10^4} < \frac{10}{10^4}\)

\(10^{-4} < \frac{ 1.25}{10^4} < 10^{-3}\)

k = -3­

Answer (D)

Method 2:

\(10^{k-1} < 0.000125 < 10^k\)

\(1*10^{k-1} < 1.25 * 10^{-4} < 10*10^{k-1}\)

Hence k - 1 = -4
k = -3
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 23 Apr 2024, 05:18
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­10^(k-1) < 1.25* 10^-4 < 10^-k 

­10^(k-1) <10^-4 < 10^-k 
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 17 May 2024, 09:59
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­Approximate that middle term a bit. No sweat:

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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 20 Jun 2024, 03:55
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Bunuel please attach more questions like these!
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 20 Jun 2024, 05:19
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aryamannm750+
Bunuel please attach more questions like these!
­
Check Exponents and Inequalities Questions.
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 16 Aug 2024, 06:28
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I solved it this way:

If you remember that 1/8 = 0.125

Then

.000125 = (1/10^3)(1/8)
or
.000125 = (10^-3)(1/8)

Because

(1/10) < (1/8) < (1)

Then we know that

(1/10^3)(1/10) < (1/10^3)(1/8) < (1/10^3)(1)

or

(1/10^3)(1/10) < .000125 < (1/10^3)(1)

Which means

(1/10^4) < .000125 < (1/10^3)

or

10^-4 < .000125 < 10^-3

Therefore, k=-3
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 04 Dec 2024, 07:04
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 26 May 2025, 08:42
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how did you come up with k+5=2
gmatophobia
Bunuel
If k is an integer and \(10^{k-1} < 0.000125 < 10^k\), then k =

A. -6
B. -5
C. -4
D. -2
E. -1

\(10^{k-1} < 0.000125 < 10^k\)

\(10^{k-1} < 125*10^{-6} < 10^k\)

Dividing all sides by \(10^{-6}\)

\(10^{k-1+6} < 125 < 10^{k+6}\)

\(10^{k+5} < 125 < 10^{k+6}\)

Therefore \(k + 5 = 2\)

\(k = -3\)
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If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = 26 May 2025, 08:50
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shubhim20
how did you come up with k+5=2
gmatophobia
Bunuel
If k is an integer and \(10^{k-1} < 0.000125 < 10^k\), then k =

A. -6
B. -5
C. -4
D. -2
E. -1

\(10^{k-1} < 0.000125 < 10^k\)

\(10^{k-1} < 125*10^{-6} < 10^k\)

Dividing all sides by \(10^{-6}\)

\(10^{k-1+6} < 125 < 10^{k+6}\)

\(10^{k+5} < 125 < 10^{k+6}\)

Therefore \(k + 5 = 2\)

\(k = -3\)

That doubt has been addressed a couple of times above.

Basically, since 125 lies between two consecutive powers of 10, we know that 100 < 125 < 1,000, or 10^2 < 125 < 10^3. That means k + 5 = 2, which gives k = -3.
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