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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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Question Stats:

88% (00:56) correct 11% (00:54) wrong based on 17 sessions


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
D,E
[Reveal] Spoiler: OA

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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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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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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?
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Re: 3^2a 11^b [#permalink] New post 08 Jul 2018, 09:51
Expert's post
Actuallly

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

AND

\(11^b = 11^{2x}\)

Hope you spot the gist of the problem.

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Re: 3^2a 11^b   [#permalink] 08 Jul 2018, 09:51
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