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GRE Math Challenge #85-Simplify 2^x + 2^x

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GRE Math Challenge #85-Simplify 2^x + 2^x [#permalink] New post 09 May 2015, 10:27
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Question Stats:

30% (00:12) correct 70% (00:28) wrong based on 10 sessions
\(2^x + 2^x\)

(A) \(2^{x+1}\)
(B) \(2^{x+2}\)
(C) \(2^{2x}\)
(D) \(4^{x}\)
(E) \(4^{2x}\)
[Reveal] Spoiler: OA

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Re: GRE Math Challenge #85-Simplify 2^x + 2^x [#permalink] New post 14 Dec 2017, 08:58
sandy wrote:
\(2^x + 2^x\)

(A) \(2^{x+1}\)
(B) \(2^{x+2}\)
(C) \(2^{2x}\)
(D) \(4^{x}\)
(E) \(4^{2x}\)


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

This can also be proved

by taking x = 3

then \(2^3 + 2^3\) =\((2^3)\) * \(2^1\) = 16 i.e \(2^{3 + 1}\)
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Re: GRE Math Challenge #85-Simplify 2^x + 2^x [#permalink] New post 14 Dec 2017, 10:18
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Expert's post
sandy wrote:
\(2^x + 2^x\)

(A) \(2^{x+1}\)
(B) \(2^{x+2}\)
(C) \(2^{2x}\)
(D) \(4^{x}\)
(E) \(4^{2x}\)


We can combine the terms in the same way we combine any like terms.
For example, k + k = 2k
And w² + w² = 2(w²)
And 3q³ + 3q³ = 2(3q³) = 6q³

Likewise, 2^x + 2^x = (2)2^x
= (2^1)(2^x)
= 2^(x + 1) [after we apply the product law]

Answer: A

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Re: GRE Math Challenge #85-Simplify 2^x + 2^x   [#permalink] 14 Dec 2017, 10:18
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GRE Math Challenge #85-Simplify 2^x + 2^x

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