8^10+4^10 8^4 + 4^11 = 2^30+2^202^12+2^22= 2^20(2^10+1)2^12(1+2^10) = 2^8 = 2^4 16

smc

prealgebra intermediate

Problem

\frac{8 ^{10} + 4 ^{10}}{8 ^{4} + 4 ^{11}} =

(A)

2

(B)

16

(C)

32

(D)

(12)^{\frac{2}{3}}

Solution

\frac{8 ^{10} + 4 ^{10}}{8 ^{4} + 4 ^{11}} = \frac{2 ^{30} + 2 ^{20}}{2 ^{12} + 2 ^{22}} = \frac{2 ^{20} ( 2 ^{10} + 1 )}{2 ^{12} ( 1 + 2 ^{10} )} = 2^{8} = 2^{4}

16

Final answer