Question 273434: I saw a problem, 8^-5/3, worked like this:
(1/8)^-5/3 = (cube root of (1/8))^5 = (1/2)^5 = 1/32
I see 8^-5/3 going to (1/8)^-5/3. Why is 1/8 raised to a power? I don't understand why it's not 1/8^5/3.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 8^-5/3, worked like this:
(1/8)^-5/3 = (cube root of (1/8))^5 = (1/2)^5 = 1/32
I see 8^-5/3 going to (1/8)^-5/3. Why is 1/8 raised to a power? I don't understand why it's not 1/8^5/3.
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I think you are right; the reciprocal gets rid of the neg on the exponent
It should be (1/8)^5/3
:
Prove this to yourself on a calc:
enter 8^(-5/3) results .03125
Reciprocal removes the neg on the exponent, but the value is the same
then enter (1/8)^(5/3) = .03125
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the denominator 3 means the cube root of 1/8 which is 1/2 (1/2 * 1/2 * 1/2 = 1/8)
then that value is raised to 1/2^5 which is 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32
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This too, you can check on your calc: 1/32 = .03124
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Did this help clear things up for you?
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Response to comment, 8^(-5/3) is NOT transformed to (1/8)^(-5/3)
It is transformed to (1/8)^(+5/8), the reciprocal will change the sign of the exponent
Think of it like this: (8/1)^(-5/3) = (1/8)^(+5/3)
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Prove it to yourself with a calculator
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