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x > 1

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x > 1 [#permalink] New post 06 Nov 2018, 14:25
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Question Stats:

75% (00:31) correct 25% (00:34) wrong based on 8 sessions
\(x > 1\)

Quantity A
Quantity B
\(x(x^2)^4\)
\((x^3)^3\)


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.
[Reveal] Spoiler: OA

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Re: x > 1 [#permalink] New post 08 Nov 2018, 03:36
Carcass wrote:
\(x > 1\)

Quantity A
Quantity B
\(x(x^2)^4\)
\((x^3)^3\)


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.



Explanation::

QTY A : \(x(x^2)^4 = x^9\)

QTY B : \((x^3)^3 = x^9\)

Hence QTY A = QTY B

Option C
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Re: x > 1 [#permalink] New post 10 Nov 2018, 09:36
Carcass wrote:
\(x > 1\)

Quantity A
Quantity B
\(x(x^2)^4\)
\((x^3)^3\)


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.


Why is it wrong that I did first did \((x^1 *x^2)\)= \((x^3)\) then \((x^3)^4\)=x^12?
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Re: x > 1 [#permalink] New post 15 Nov 2018, 04:56
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Mercychee wrote:

Why is it wrong that I did first did \((x^1 *x^2)\)= \((x^3)\) then \((x^3)^4\)=x^12?


BODMAS/PEMDAS: Solve brackets or parentheses first. The 4 in \((x^2)^4\) is a part of the bracket that requires first priority while solving.
Re: x > 1   [#permalink] 15 Nov 2018, 04:56
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