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(2^5)^4+(2^3)^2=

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(2^5)^4+(2^3)^2= [#permalink]

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New post 16 Aug 2017, 00:28
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Question Stats:

85% (00:32) correct 15% (00:04) wrong based on 51 sessions

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Re: (2^5)^4+(2^3)^2= [#permalink]

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New post 16 Aug 2017, 00:46
Bunuel wrote:
\((2^5)^4+(2^3)^2=\)


A. \(2^{15}\)

B. \(2^{20}\)

C. \(2^{26}\)

D. \(2^6*(2^{14}+1)\)

E. \(2^{120}*(2^2+1)\)


\((2^5)^4+(2^3)^2=2^{20}+2^6=2^6(2^{14}+1)\)

Answer D
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Re: (2^5)^4+(2^3)^2= [#permalink]

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New post 16 Aug 2017, 07:23
Bunuel wrote:
\((2^5)^4+(2^3)^2=\)


A. \(2^{15}\)

B. \(2^{20}\)

C. \(2^{26}\)

D. \(2^6*(2^{14}+1)\)

E. \(2^{120}*(2^2+1)\)

\((2^5)^4+(2^3)^2\)

\(2^{20}+2^6\)

\(2^6(2^{14}+1)\)

Answer (D)...

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(2^5)^4+(2^3)^2= [#permalink]

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New post 17 Aug 2017, 13:20
Bunuel wrote:
\((2^5)^4+(2^3)^2=\)


A. \(2^{15}\)

B. \(2^{20}\)

C. \(2^{26}\)

D. \(2^6*(2^{14}+1)\)

E. \(2^{120}*(2^2+1)\)


(2^5)^4 = 2^20 + 2^6
-- When powers are set to another power, you multiply them together

2^6(2^14+1)
-- When adding powers together x^y + z^y, you have to factor out the smallest power, and multiply by that same number; in this case, 2^6 < 2^20, so you factor out with 2^6 and then multiply by 2^6
20 - 6 = 14
6 -6 = 0 (anything to the power of 0 turns into 1)
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(2^5)^4+(2^3)^2=   [#permalink] 17 Aug 2017, 13:20
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