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a > 1 and b > 5
[#permalink]
18 Jul 2018, 11:25
18 Jul 2018, 11:25
Expert Reply
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Question Stats:
77% (01:01) correct
22% (01:30) wrong based on 130 sessions
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\(a > 1\)
\(b > 5\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Kudos for R.A.E
\(b > 5\)
Quantity A |
Quantity B |
\((5b)^a\) |
\((b^2)^a\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Kudos for R.A.E
Re: a > 1 and b > 5
[#permalink]
19 Jul 2018, 01:35
19 Jul 2018, 01:35
1
Let us expand both quantities.
Starting with qty A
\((5b)^a\) can be re-written as \(5^a * b^a\)
qty B is equal to
\(b^2^a\)
Let us divide both sides by \(b^a\); remember that \(b\) is positive so when be is raised to any power the result will be +ve.
qty A will equal
\(5^a\)
while qty B will equal
\(b^a\)
Since \(b > 5\) ; \(b^a\) will always be greater than \(5^a\) since a is also a +ve no greater than 1
Starting with qty A
\((5b)^a\) can be re-written as \(5^a * b^a\)
qty B is equal to
\(b^2^a\)
Let us divide both sides by \(b^a\); remember that \(b\) is positive so when be is raised to any power the result will be +ve.
qty A will equal
\(5^a\)
while qty B will equal
\(b^a\)
Since \(b > 5\) ; \(b^a\) will always be greater than \(5^a\) since a is also a +ve no greater than 1
Re: a > 1 and b > 5
[#permalink]
24 Nov 2019, 09:50
24 Nov 2019, 09:50
amorphous wrote:
Let us expand both quantities.
Starting with qty A
\((5b)^a\) can be re-written as \(5^a * b^a\)
qty B is equal to
\(b^2^a\)
Let us divide both sides by \(b^a\); remember that \(b\) is positive so when be is raised to any power the result will be +ve.
qty A will equal
\(5^a\)
while qty B will equal
\(b^a\)
Since \(b > 5\) ; \(b^a\) will always be greater than \(5^a\) since a is also a +ve no greater than 1
Starting with qty A
\((5b)^a\) can be re-written as \(5^a * b^a\)
qty B is equal to
\(b^2^a\)
Let us divide both sides by \(b^a\); remember that \(b\) is positive so when be is raised to any power the result will be +ve.
qty A will equal
\(5^a\)
while qty B will equal
\(b^a\)
Since \(b > 5\) ; \(b^a\) will always be greater than \(5^a\) since a is also a +ve no greater than 1
How do you get b^a after dividing b^2a by b^a. It should be b^2.
I think the answer is D.
