Question

can someone please help me? Medal awarded xx please!!
evaluate no calculators
81^-3/4

(9/4)^-5/2

(1000/343)^4/3

Answer 1

instead of (1000/343) the other problem is
(27/1000)^-2/3

Answer 2

\[81^{-3/4} = \frac{1}{81^{3/4}}\]by \[a^{-n} = \frac{1}{a^n}\]then by rules for fractional exponents, \[(x^a)^b = x^{a*b}\]so\[\frac{1}{81^{3/4}} = \frac{1}{(81^{1/4})^3}= \frac{1}{(\sqrt[4]{81})^3}\]
Can you find the 4th root of 81?

Answer 3

Hint:\[\sqrt[4]{x} = \sqrt{\sqrt{x}}\]

Answer 4

3

Answer 5

okay, so that leaves us with \[\frac{1}{(\sqrt[4]{81})^3} = \frac{1}{3^3} =\]

Answer 6

1/9

Answer 7

look more closely

Answer 8

oh 27 haha

Answer 9

much better. the others should be very similar...

Answer 10

so looking at (27/1000)^-2/3, would i have to multiply by 3/2?

Answer 11

\[(\frac{27}{1000})^{-2/3} = (\frac{1000 } {27})^{2/3}\]

Answer 12

would the answer be 100/9?

Answer 13

It would indeed.

Answer 14

thank you soooo much! That makes so much sense now. Thank you for your time xx

Answer 15

very easy to compute those if you realize that \(1000 = 10^3\) and \(27 = 3^3\), then it's just \[(\frac{10^3}{3^3})^{2/3} = \frac{10^{3*(2/3)}}{3^{3*(2/3)}}=\frac{10^2}{3^2}\]

Answer 16

yes so for the (9/4)^-5/2 would that be the same concept?

Answer 17

you could do it that way, yes. make the numbers jump! :-)

Answer 18

:D hahahaha so basically it would be 4/9^5/2 format then solve it from there

Answer 19

yes. or if you're doing it that last way, you could just wait until the end to flip it over. 6 of one, half dozen of the other...

Answer 20

it would equal 32/243

Answer 21

time for me to call it a night. looks like you have this well under control!

Answer 22

thanks again so much! :) you are the best :D