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If y = 2^(x+1), what is the value of y – x?

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If y = 2^(x+1), what is the value of y – x?  [#permalink]

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New post Updated on: 10 Aug 2017, 01:45
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If \(y = 2^{(x+1)}\), what is the value of y – x?


(1) \(2^{(2x+2)} = 64\)

(2) \(y = 2^{(2x –1)}\)

Originally posted by Lolaergasheva on 05 Mar 2011, 08:12.
Last edited by Bunuel on 10 Aug 2017, 01:45, edited 2 times in total.
Edited the question.
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Re: If y = 2^(x+1), what is the value of y – x?  [#permalink]

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New post Updated on: 05 Mar 2011, 09:40
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Lolaergasheva wrote:
If y = 2^x+1, what is the value of y – x?
(1) 2^2x+2 = 64
(2) y = 2^2x –1

1) 2^2x+2=2^6
2x+2=6
x=2
2^x+1-2^1 => x+1-1=0 ==> x=0

2) I don't know how to solve


I assume that the questions reads as follows:

If y = \(2^{x+1}\), what is the value of y – x?
(1) \(2^{2x+2}\) = 64
(2) \(y = 2^{2x-1}\)


Then, 1st statement gives us 2x+2 = 6 and hence x =2 and therefore y = \(2^{2+1}\) = \(2^3\) = 8 hence y-x = 8-2 = 6, So sufficient

From 2, \(y = 2^{2x-1}\) = \(2^{x+1}\)

or 2x-1 = x+1 and hence x =2 and hence like in 1, y =8 and hence y-x = 8-2 = 6, So sufficient

Answer is D

Originally posted by beyondgmatscore on 05 Mar 2011, 08:24.
Last edited by beyondgmatscore on 05 Mar 2011, 09:40, edited 2 times in total.
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Re: If y = 2^(x+1), what is the value of y – x?  [#permalink]

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New post 05 Mar 2011, 08:45
Lolaergasheva wrote:
If y = 2^x+1, what is the value of y – x?
(1) 2^2x+2 = 64
(2) y = 2^2x –1


1) 2^2x+2=2^6
2x+2=6
x=2
2^x+1-2^1 => x+1-1=0 ==> x=0

2) I don't know how to solve


Please, format the questions properly.

Original question is:
If y = 2^(x+1), what is the value of y – x?

(1) 2^(2x+2) = 64 --> \(2^{2x+2}=2^6\) --> as bases are equal we can equate the powers: \(2x+2=6\) --> \(x=2\). From the stem: \(y=2^{x+1}=2^3=8\), so \(y-x=8-2=6\). Sufficient.

(2) y = 2^(2x –1) --> as also given that \(y=2^{x+1}\) then \(2^{x+1}=2^{2x-1}\) --> \(x+1=2x-1\) --> \(x=2\), then the same way as above \(y=8\) --> \(y-x=8-2=6\). Sufficient.

Answer: D.
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Re: If y = 2^(x+1), what is the value of y – x?  [#permalink]

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New post 10 Nov 2015, 11:11
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If y = 2^(x+1), what is the value of y – x?

(1) 2^(2x+2) = 64

(2) y = 2^(2x –1)

There are 2 variables (x,y) one equation (y = 2^(x+1)), and 2 more equations from the 2 conditions, so there is high chance (D) will be our answer.
From condition 1, 2^(2x+2)=64=2^6, or 2x+2=6, x=2, y=2^3=8. This is sufficient
From condition 2, 2^(2x-1)=2^(x+1), or 2x-1=x+1, x=2, y=8 . This is sufficient as well.
Therefore, the answer becomes (D).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If y = 2^(x+1), what is the value of y – x?  [#permalink]

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New post 21 Dec 2016, 00:09
(1) 2^2x+2=64
2^2x+2=2^6
2x+2=6
x=2, y=2^x+1=2^2+1=8 Sufficient
(2) y=2^2x-1
2^2x-1=2^x+1
2x-1=x+1
x=2, y=8 Sufficient

Answer D
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Re: If y = 2^(x+1), what is the value of y – x?  [#permalink]

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New post 16 Jul 2017, 15:04
If \(y = 2^{x+1}\), what is the value of y – x?

(1) \(2^{2x+2} = 64\)

\(2^{2x+2} = 2^6\)

\(2x+2 = 6\)

\(2x = 4\)

\(x = 2\)

\(y = 2^{x+1}\)

\(y = 2^{2+1} = 2^3 = 8\)

\(y - x = 8 - 2 = 6\)

Hence, (1) ===== is SUFFICIENT

(2) \(y = 2^{2x –1}\)

\(y = 2^{2x –1}\)

\(y = 2^{x+1}\)

\(2^{2x –1} = = 2^{x+1}\)

\(2x - 1 = x + 1\)

\(x = 2\)

\(y = 2^{x+1}\)

\(y = 2^{2+1} = 2^3 = 8\)

\(y - x = 8 - 2 = 6\)

Hence, (2) ===== is SUFFICIENT

Hence, Answer is D

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Re: If y = 2^(x+1), what is the value of y – x?  [#permalink]

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New post 22 Jul 2018, 14:47
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