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If m > 1 and n = 2^(m 1), then 4^m =

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Bunuel
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If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   03 Dec 2014, 08:28
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A
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Question Stats:

78% (01:52) correct 22% (02:14) wrong based on 348 sessions
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If m > 1 and n = 2^(m −1), then 4^m =

A) 16n^2
B) 4n^2
C) n^2
D) n^2/4
E) n^2/16



Source: Chili Hot GMAT
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B
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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   03 Dec 2014, 08:40
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n = 2^(m −1)
=> n = 2^m/ 2
=> 2n = 2^m

Squaring both sides;
4n^2 = 4^m

Hence B
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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   03 Dec 2014, 09:39
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2^m = 2n.
4^m = 2^(2m) = (2^m)^2
=> (2n)^2 = 4n^2

Answer is B.
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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   03 Dec 2014, 20:42
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Answer = B) 4n^2

\(n = 2^{m-1}\)

\(n = \frac{2^m}{2}\)

\(2^m = 2n\)

Squaring both sides

\({(2^m)}^{2} = (2n)^2\)

\(2^{2m} = 4n^2\)

\(4^m = 4n^2\)
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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   04 Dec 2014, 00:13
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If m > 1 and n = 2^(m −1), then 4^m
Given m>1, so let's assume m=2
n=2^(m-1) = 2^(2-1) = 2, so n=2

hence, 4^m = 4^2 = 16
only 1 ans. choice can satisfy this:
A) 16n^2 --> clearly > 16
B) 4n^2 --> 4*2^2 = 16 (we can stop after this as there can be only 1 right answer)
C) n^2 --> clearly < 16
D) n^2/4 --> clearly < 16
E) n^2/16 --> clearly < 16

Ans. B) 4n^2
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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   04 Dec 2014, 07:26
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Bunuel wrote:

Tough and Tricky questions: Exponents.



If m > 1 and n = 2^(m −1), then 4^m =

A) 16n^2
B) 4n^2
C) n^2
D) n^2/4
E) n^2/16

Kudos for a correct solution.

Source: Chili Hot GMAT


The correct answer is B.
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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   08 Mar 2015, 13:33
PareshGmat wrote:
Answer = B) 4n^2

\(n = 2^{m-1}\)

\(n = \frac{2^m}{2}\)

These steps I don't understand, could you please explain?

Why is it that you divide by 2.
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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   08 Mar 2015, 13:35
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erikvm wrote:
PareshGmat wrote:
Answer = B) 4n^2

\(n = 2^{m-1}\)

\(n = \frac{2^m}{2}\)

These steps I don't understand, could you please explain?

Why is it that you divide by 2.


This is basic staff: \(2^{m-1}=2^m*2^{-1}=\frac{2^m}{2}\).
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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   08 Mar 2015, 13:40
Ah thats right. Jeez, when something is slightly disguised, I get so thrown off.

It's because \(2^m\) * 2^-1 = 2^m * 1/2, correct?
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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   08 Mar 2015, 13:43
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erikvm wrote:
Ah thats right. Jeez, when something is slightly disguised, I get so thrown off.

It's because \(2^m\) * 2^-1 = 2^m * 1/2, correct?

______
Yes, 2^(-1) = 1/2.
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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   08 Mar 2015, 22:34
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Hi All,

This question is perfect for TESTing VALUES....

We're told that M > 1 and N = 2^(M-1). We're asked for the value of 4^M

IF.....
M = 2
N = 2^(1) = 2
The value of 4^M = 4^2 = 16

So we're looking for an answer that = 16 when N = 2

Answer A: 16(2^2) = 64 This is NOT a match
Answer B: 4(2^2) = 16 This IS a match
Answer C: 2^2 = 4 This is NOT a match
Answer D: (2^2)/4 = 1 This is NOT a match
Answer E: (2^2)/16 = 1/4 This is NOT a match

Final Answer:
Show Spoiler
B


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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   08 Mar 2016, 10:21
Bunuel wrote:

Tough and Tricky questions: Exponents.



If m > 1 and n = 2^(m −1), then 4^m =

A) 16n^2
B) 4n^2
C) n^2
D) n^2/4
E) n^2/16

Kudos for a correct solution.

Source: Chili Hot GMAT



Excellent Question
Here 2^m = 2 x n
now 4^m = 2^2m= (2^m)^2= 4n^2
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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   06 Jul 2020, 13:47
\(n = 2^{(m −1)}\)
\(n=2^m*2^{-1}\)
\(n=\frac{2^m}{2}\)
\(2n=2^m\)
\(4^m=4n^2\)
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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink] New post   30 Jul 2023, 05:58
n = 2^(m −1)
Simplifying,
=>n = 2^(m −1) = (2^m)/2
=>2n = (2^m)
Or =>(2^m) = 2n
Squaring both the sides
=>(2^m)^2 = (2n)^2
=> 4^m=4n^2
Hence B
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Re: If m > 1 and n = 2^(m 1), then 4^m = [#permalink]
30 Jul 2023, 05:58
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