How do I evaluate the following two expressions. Write your answer without exponents.
32 ^ -4/5 =
(1 / 27) ^ 2/3 =
2007-05-20 21:12:36 · 6 answers · asked by ♥Cheerleader♥ 1 in Science & Mathematics ➔ Mathematics
You need to know three things
a........The negative index means 1/ something
b.........The denominator of 5 means the fifth root
c.........The denominator of 3 means the cube root
so
Q1
32 ^ -4/5
= (1/32)^4/5... this is the 1/ something
= (1/2) ^ 4 .... the fifth root of 32 is 2
= 1/16
Q2
(1 / 27) ^ 2/3
= (1/3)^2 ....... the cube root of 27 is 3
= 1/9
2007-05-20 22:25:34 · answer #1 · answered by fred 5 · 0⤊ 0⤋
Question 1
= 1 / 32^(4/5)
= 1 / 2^4
= 1 / 16
Question 2
= 1 / 3 ²
= 1 / 9
2007-05-21 05:22:52 · answer #2 · answered by Como 7 · 0⤊ 0⤋
32 ^ -4/5 = 1/(32)^(4/5)
= 1/ 2^4
= 1/16
(1 / 27) ^ 2/3 = (1/3)^2 = 1/9
2007-05-21 04:18:57 · answer #3 · answered by michael_scoffield 3 · 0⤊ 0⤋
32 ^ -4/5 =
= 2^[(5)(-4/5)]
= 2^-4
= 1/16
(1 / 27) ^ 2/3 =
= [1/(3^3)] ^ 2/3
= (3^-3)^2/3
= 3^[(-3)(2/3)]
= 3^-2
=1/9
2007-05-21 04:24:53 · answer #4 · answered by misshahila 2 · 0⤊ 0⤋
its by using the law of indices ...
a^-m = 1/a^m..
so 32^-4/5 = 1/32^4/5 =1/16
and (1/27)^2/3 = 1/9
2007-05-21 04:23:49 · answer #5 · answered by Harish R 2 · 0⤊ 0⤋
rewrite:
1/(5th root(32))^4
2^5 = 32
(1/(2)^4) = 1/16
1/(3rd root(27)^2)
3^3 = 27
1/(3^2) = 1/9
2007-05-21 04:22:30 · answer #6 · answered by wilmer 5 · 0⤊ 0⤋