what is the reciprocal for
(81/16)^-3/4
2007-10-13 04:04:29 · 4 answers · asked by Cool_Q 4 in Science & Mathematics ➔ Mathematics
kindly note the one who answered above that 3^3 is 27, and not 9. here's a more detailed solution...
(81/16)^-3/4
based from the laws of exponents, you may also write this as
(81/16)^-3/4 = [(81/16)^1/4]^-3
taking the 4th root of both 81 and 16, we have 3 and 2 respectively, because
3*3*3*3 = 81 and 2*2*2*2 = 16, thus,
(81/16)^-3/4 = [3/2]^-3
again, from the laws of exponents,
a^(-x) = (1/a)^x, and so...
(81/16)^-3/4 = [3/2]^-3
= [2/3]^3
so take the cube of 2 and 3, we have 8 and 27 respectively because,
2*2*2 = 8 and 3*3*3=27
therefore
(81/16)^-3/4 = [2/3]^3 = 8/27 so
(81/16)^-3/4 = 8/27
2007-10-13 04:18:10 · answer #1 · answered by tootoot 3 · 0⤊ 1⤋
(81/16)^-3/4 = (3^4/2^4)^-3/4 = [(3/2)^4]^-3/4 =(3/2) ^-3
reciprocal is (3/2)^3 = 27/8
2007-10-13 11:26:20 · answer #2 · answered by pereira a 3 · 0⤊ 0⤋
(81/16)^-3/4
= (16/81)^3/4
= [ (16/81)^1/4 ]^3
= [ 16^1/4 / 81^1/4 ]^3
= [2/3]^3
= 2^3 / 3^3
= 8/9
,.,.,.,.,.,..,...
2007-10-13 11:13:24 · answer #3 · answered by The Wolf 6 · 2⤊ 1⤋
Oops 3^3 = 27
so the answer should be 8/27
2007-10-13 11:18:20 · answer #4 · answered by norman 7 · 0⤊ 0⤋