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Which is greater 7^1/x-1/y or 7^x-y
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16 Apr 2019, 00:35
16 Apr 2019, 00:35
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33% (01:15) wrong based on 30 sessions
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\(0 < x < y\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(7^{\frac{1}{x}-\frac{1}{y}}\) |
\(7^{x-y}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Which is greater 7^1/x-1/y or 7^x-y
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17 Apr 2019, 05:25
17 Apr 2019, 05:25
1
Carcass wrote:
\(0 < x < y\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(7^{\frac{1}{x}-\frac{1}{y}}\) |
\(7^{x-y}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
QUANTITY A:
\(\frac{1}{x}-\frac{1}{y}=\frac{y}{xy}-\frac{x}{xy}\)
\(=\frac{y-x}{xy}\)
Since 0< x < y, we know that xy is POSITIVE, and we know that y-x is POSITIVE
So, \(\frac{y-x}{xy}=\frac{POSITIVE}{POSITIVE}=POSITIVE\)
So, \(7^{\frac{1}{x}-\frac{1}{y}}=7^{POSITIVE}\)
= some value greater than 1
QUANTITY B:
Since 0< x < y, we know that \(x-y\) is NEGATIVE
So, \(7^{x-y}=7^{NEGATIVE}\)
= some value less than 1
Answer: A
Cheers,
Brent
Re: Which is greater 7^1/x-1/y or 7^x-y
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07 Mar 2022, 04:50
07 Mar 2022, 04:50
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