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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D-Day: November 18th, 2017

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