Complicated exponent

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Complicated exponent

by GeauxSwish » Tue Feb 07, 2017 5:17 pm
Problem: (x^-5/x^-9)^1/2.

The answer is x2, but here is my confusion


To do this problem my first instinct was to just flip the fraction in the parenthesis, but someone I've been working with told me that's not possible. There approach was to multiply the denominator by a negative fraction, and then multiply. But I'm still confused as to why his version is right and mine is wrong

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by Brent@GMATPrepNow » Tue Feb 07, 2017 5:42 pm
GeauxSwish wrote:Problem: (x^-5/x^-9)^1/2.

The answer is x2, but here is my confusion


To do this problem my first instinct was to just flip the fraction in the parenthesis, but someone I've been working with told me that's not possible. There approach was to multiply the denominator by a negative fraction, and then multiply. But I'm still confused as to why his version is right and mine is wrong
IMPORTANT RULES
Division Rule: (x^a)/(x^b) = x^(a-b)
Power of a Power Rule: (x^a)^b = x^(ab)

(x^-5/x^-9)^1/2 = (x^(-5 - -9))^1/2 [Division Rule]
= (x^4)^1/2
= x^(4 times 1/2) [Power of a Power Rule]
= x^2

Cheers,
Brent
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by GeauxSwish » Tue Feb 07, 2017 6:09 pm
Brent@GMATPrepNow wrote:
GeauxSwish wrote:Problem: (x^-5/x^-9)^1/2.

The answer is x2, but here is my confusion


To do this problem my first instinct was to just flip the fraction in the parenthesis, but someone I've been working with told me that's not possible. There approach was to multiply the denominator by a negative fraction, and then multiply. But I'm still confused as to why his version is right and mine is wrong
IMPORTANT RULES
Division Rule: (x^a)/(x^b) = x^(a-b)
Power of a Power Rule: (x^a)^b = x^(ab)

(x^-5/x^-9)^1/2 = (x^(-5 - -9))^1/2 [Division Rule]
= (x^4)^1/2
= x^(4 times 1/2) [Power of a Power Rule]
= x^2

Cheers,
Brent
So if I understand correct my way where I just flip the whole fraction so its x^9/x5 is just straight up illegal

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by Brent@GMATPrepNow » Tue Feb 07, 2017 7:09 pm
GeauxSwish wrote:
Brent@GMATPrepNow wrote:
GeauxSwish wrote:Problem: (x^-5/x^-9)^1/2.

The answer is x2, but here is my confusion


To do this problem my first instinct was to just flip the fraction in the parenthesis, but someone I've been working with told me that's not possible. There approach was to multiply the denominator by a negative fraction, and then multiply. But I'm still confused as to why his version is right and mine is wrong
IMPORTANT RULES
Division Rule: (x^a)/(x^b) = x^(a-b)
Power of a Power Rule: (x^a)^b = x^(ab)

(x^-5/x^-9)^1/2 = (x^(-5 - -9))^1/2 [Division Rule]
= (x^4)^1/2
= x^(4 times 1/2) [Power of a Power Rule]
= x^2

Cheers,
Brent
So if I understand correct my way where I just flip the whole fraction so its x^9/x^5 is just straight up illegal
Your way is correct.
x^(-5) = 1/(x^5) and x^(-9) = 1/(x^9)

So, (x^-5/x^-9) = [1/(x^5)]/[1/(x^9)]
= [1/(x^5)]/[(x^9)/1]
= x^9/x^5
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by Matt@VeritasPrep » Fri Feb 17, 2017 2:04 am
GeauxSwish wrote: So if I understand correct my way where I just flip the whole fraction so its x^9/x5 is just straight up illegal
That's actually fine!

You should get x� / x�, which is the same thing as x�. All that's left is to take the square root of x� (since you're raising to the 1/2 power, which is the same thing as taking the root), and you're done!