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# If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k =

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Re: If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = [#permalink]
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Bunuel wrote:
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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Re: If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = [#permalink]
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Bunuel wrote:
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

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Re: If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = [#permalink]
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gmatophobia wrote:
Bunuel wrote:
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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Re: If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = [#permalink]
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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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Re: If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = [#permalink]
gmatophobia wrote:
Bunuel wrote:
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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Re: If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = [#permalink]
Hoozan wrote:
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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Re: If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = [#permalink]
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Bunuel wrote:
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

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Re: If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = [#permalink]
­10^(k-1) < 1.25* 10^-4 < 10^-k

­10^(k-1) <10^-4 < 10^-k
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Re: If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = [#permalink]
­Approximate that middle term a bit. No sweat:

Re: If k is an integer and 10^(k - 1) < 0.000125 < 10^k, then k = [#permalink]
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