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

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Math Expert
Joined: 02 Sep 2009
Posts: 54496

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16 Aug 2017, 00:28
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Difficulty:

15% (low)

Question Stats:

86% (00:46) correct 14% (00:22) wrong based on 71 sessions

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

A. $$2^{15}$$

B. $$2^{20}$$

C. $$2^{26}$$

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

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

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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. $$2^{120}*(2^2+1)$$

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

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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. $$2^{120}*(2^2+1)$$

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

$$2^{20}+2^6$$

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

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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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