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3^2a 11^b

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3^2a 11^b [#permalink] New post 07 Aug 2017, 03:27
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81% (01:13) correct 18% (01:35) wrong based on 77 sessions
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Director
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Re: 3^2a 11^b [#permalink] New post 24 Sep 2017, 07:01
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Here there is a missing OA!
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VP
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Re: 3^2a 11^b [#permalink] New post 24 Sep 2017, 07:32
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Carcass wrote:
If \(3^{2a}\)\(11^b\)= \(27^{4x}\) \(33^{2x}\) then x must equal which of the following ?

Indicate all that apply.

❑ 2a

❑ 2b

❑ 7a - 2b

❑ \(\frac{a}{7}\)

❑ \(\frac{b}{2}\)

[Reveal] Spoiler: OA


Here given

\(3^{2a}\)\(11^b\)= \(27^{4x}\) \(33^{2x}\)


\(3^{2a}\)\(11^b\) = \(3^{12x}\) \(3^{2x}\) \(11^{2x}\) (Since \(27^{4x}\) = \({3^{(3x)}}^{4}\) = \(3^{12x}\))

or \(3^{2a}\)\(11^b\) = \(3^{14x}\) \(11^{2x}\)

Since prime bases are same, the exponents must also be equal.
14x = 2a,

or x= \(\frac{2}{14}\)

or a =\(\frac{a}{7}\)

And 2x = b, or x= \(\frac{b}{2}\)

Therefore only choices (D) and (E) must be true
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Intern
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Re: 3^2a 11^b [#permalink] New post 08 Jul 2018, 08:19
pranab01 wrote:
Carcass wrote:
If \(3^{2a}\)\(11^b\)= \(27^{4x}\) \(33^{2x}\) then x must equal which of the following ?

Indicate all that apply.

❑ 2a

❑ 2b

❑ 7a - 2b

❑ \(\frac{a}{7}\)

❑ \(\frac{b}{2}\)

[Reveal] Spoiler: OA


Here given

\(3^{2a}\)\(11^b\)= \(27^{4x}\) \(33^{2x}\)


\(3^{2a}\)\(11^b\) = \(3^{12x}\) \(3^{2x}\) \(11^{2x}\) (Since \(27^{4x}\) = \({3^{(3x)}}^{4}\) = \(3^{12x}\))

or \(3^{2a}\)\(11^b\) = \(3^{14x}\) \(11^{2x}\)

Since prime bases are same, the exponents must also be equal.
14x = 2a,

or x= \(\frac{2}{14}\)

or a =\(\frac{a}{7}\)

And 2x = b, or x= \(\frac{b}{2}\)

Therefore only choices (D) and (E) must be true


I know that one base equal to another will have the same exponent, but here we have two different bases. How that could be true to equate them as you did?
Founder
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Joined: 18 Apr 2015
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Re: 3^2a 11^b [#permalink] New post 08 Jul 2018, 09:51
Expert's post
Re: 3^2a 11^b   [#permalink] 08 Jul 2018, 09:51
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