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Which of the following is equal to 2^{k-1}3^{k+1}?

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Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 08 May 2018, 02:52
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[GMAT math practice question]

Which of the following is equal to \(2^{k-1}3^{k+1}\)?

\(A. 2^2(6^{k-1})\)
\(B. 3^2(6^{k-1})\)
\(C. 6(2^{k-1})\)
\(D. 6(3^{k-1})\)
\(E. 6^{k-1}\)

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Re: Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 08 May 2018, 03:21
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MathRevolution wrote:
[GMAT math practice question]

Which of the following is equal to \(2^{k-1}3^{k+1}\)?

\(A. 2^2(6^{k-1})\)
\(B. 3^2(6^{k-1})\)
\(C. 6(2^{k-1})\)
\(D. 6(3^{k-1})\)
\(E. 6^{k-1}\)


Approach 01 : From Q-stem to Answer: \(2^{k-1}3^{k+1}\) = \(3^2*(2^{k-1}3^{k-1})\) = \(3^2(6^{k-1})\) .......... Option B is the answer

Approach 02 : From Answer to Q-stem:

Option 01:\(A. 2^2(6^{k-1})\) = \(2^2*(2^{k-1}*3^{k-1})\) = \(2^{k+1}*3^{k-1}\)
Option 02:\(3^2(6^{k-1})\) = \(3^2*(2^{k-1}*3^{k-1})\) = \(2^{k-1}*3^{k+1}\) .......... Option B is the answer
Option 03:\(6(2^{k-1})\) = \(3*2*(2^{k-1})\) = \(3*2^k\)
Option 04:\(6(3^{k-1})\) = \(3*2*(3^{k-1})\) = \(2*3^k\)
Option 05:\(6^{k-1}\) = \(2^{k-1}*3^{k-1}\)
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Re: Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 08 May 2018, 03:24
MathRevolution wrote:
[GMAT math practice question]

Which of the following is equal to \(2^{k-1}3^{k+1}\)?

\(A. 2^2(6^{k-1})\)
\(B. 3^2(6^{k-1})\)
\(C. 6(2^{k-1})\)
\(D. 6(3^{k-1})\)
\(E. 6^{k-1}\)


\(2^{k-1}3^{k+1}\) \(=\) \(2^{k-1}*3^{k-1}*3^2\) \(=\) \(3^2*6^{k-1}\)

Answer: B
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Re: Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 10 May 2018, 07:58
=>

\(2^{k-1}3^{k+1} = 2^{k-1}3^{k-1}*3^2= 3^2*(2*3)^{k-1} = 3^2(6^{k-1})\)

Therefore, the answer is B.
Answer: B
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Re: Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 10 May 2018, 09:01
MathRevolution wrote:
=>

\(2^{k-1}3^{k+1} = 2^{k-1}3^{k-1}*3^2= 3^2*(2*3)^{k-1} = 3^2(6^{k-1})\)

Therefore, the answer is B.
Answer: B


how did you solve for 3^k-1 * 3^2 ??
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Re: Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 23 May 2018, 21:33
SandhyAvinash wrote:
MathRevolution wrote:
=>

\(2^{k-1}3^{k+1} = 2^{k-1}3^{k-1}*3^2= 3^2*(2*3)^{k-1} = 3^2(6^{k-1})\)

Therefore, the answer is B.
Answer: B


how did you solve for 3^k-1 * 3^2 ??


Solving this

2^(k−1)*3^(k+1)
=separate out power of 2 and 3 from k power

= 2^k*3^k*3/2
=(2*3)^k*(3/2)
=6^k *1.5

When you simplify B it will give same as

= 9(6^k)/6
=3*(6^k)/2

=1.5*(6)^k

Give kudos if it helps

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Re: Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 03 Jun 2018, 05:26
MathRevolution wrote:
=>

\(2^{k-1}3^{k+1} = 2^{k-1}3^{k-1}*3^2= 3^2*(2*3)^{k-1} = 3^2(6^{k-1})\)

Therefore, the answer is B.
Answer: B


pushpitkc any idea where from do we get additional \(3^2\) here \(2^{k-1}3^{k-1}*3^2\) ?

thank you anf happy Sunday :)
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Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 03 Jun 2018, 06:25
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dave13 wrote:
MathRevolution wrote:
=>

\(2^{k-1}3^{k+1} = 2^{k-1}3^{k-1}*3^2= 3^2*(2*3)^{k-1} = 3^2(6^{k-1})\)

Therefore, the answer is B.
Answer: B


pushpitkc any idea where from do we get additional \(3^2\) here \(2^{k-1}3^{k-1}*3^2\) ?

thank you anf happy Sunday :)



Hey dave13

\(2^{k-1}*3^{k+1} = 2^{k-1}*3^{k+1}*\frac{3^2}{3^2} = 2^{k-1}*3^{k+1}*3^{-2}*3^2\)

\(3^{k+1}*3^{-2}\) is of form \(a^m * a^n\) which is equal to \(a^{m+n}\). Here, we get \(3^{k+1-2} = 3^{k-1}\)

This further simplifies as follows: \(2^{k-1}3^{k-1}*3^2= 3^2*(2*3)^{k-1} = 3^2(6^{k-1})\)

Hope this helps you!
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Re: Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 03 Jun 2018, 07:15
pushpitkc wrote:
dave13 wrote:
MathRevolution wrote:
=>

\(2^{k-1}3^{k+1} = 2^{k-1}3^{k-1}*3^2= 3^2*(2*3)^{k-1} = 3^2(6^{k-1})\)

Therefore, the answer is B.
Answer: B


pushpitkc any idea where from do we get additional \(3^2\) here \(2^{k-1}3^{k-1}*3^2\) ?

thank you anf happy Sunday :)



Hey dave13

\(2^{k-1}*3^{k+1} = 2^{k-1}*3^{k+1}*\frac{3^2}{3^2} = 2^{k-1}*3^{k+1}*3^{-2}*3^2\)

\(3^{k+1}*3^{-2}\) is of form \(a^m * a^n\) which is equal to \(a^{m+n}\). Here, we get \(3^{k+1-2} = 3^{k-1}\)

This further simplifies as follows: \(2^{k-1}3^{k-1}*3^2= 3^2*(2*3)^{k-1} = 3^2(6^{k-1})\)

Hope this helps you!



pushpitkc many thanks for explanation just i still have question where from do you get this :? :) \(\frac{3^2}{3^2}\)

in the initial question \(2^{k-1}*3^{k+1}\) there is no such number :?
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Re: Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 03 Jun 2018, 07:23
dave13 wrote:

pushpitkc many thanks for explanation just i still have question where from do you get this :? :) \(\frac{3^2}{3^2}\)

in the initial question \(2^{k-1}*3^{k+1}\) there is no such number :?


Hey dave13

Any number multiplied by 1 is the number itself. Also, 1 can be written \(\frac{9}{9}\) which is \(\frac{3^2}{3^2}\).

That is the reason for multiplying the initial equation \(2^{k-1}*3^{k+1}\) with \(\frac{3^2}{3^2}\)

Hope this helps you!
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Re: Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 03 Jun 2018, 11:19
pushpitkc wrote:
dave13 wrote:

pushpitkc many thanks for explanation just i still have question where from do you get this :? :) \(\frac{3^2}{3^2}\)

in the initial question \(2^{k-1}*3^{k+1}\) there is no such number :?


Hey dave13

Any number multiplied by 1 is the number itself. Also, 1 can be written \(\frac{9}{9}\) which is \(\frac{3^2}{3^2}\).

That is the reason for multiplying the initial equation \(2^{k-1}*3^{k+1}\) with \(\frac{3^2}{3^2}\)

Hope this helps you!



hey pushpitkc , thank you, yes it helps :-)

can you please explain why do we need to multiply \(2^{k-1}*3^{k+1}\) with \(\frac{3^2}{3^2}\) ? what is the reason ? logic?

thank you and have a great evening :)
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Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 03 Jun 2018, 11:53
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dave13 wrote:
pushpitkc wrote:
dave13 wrote:

pushpitkc many thanks for explanation just i still have question where from do you get this :? :) \(\frac{3^2}{3^2}\)

in the initial question \(2^{k-1}*3^{k+1}\) there is no such number :?


Hey dave13

Any number multiplied by 1 is the number itself. Also, 1 can be written \(\frac{9}{9}\) which is \(\frac{3^2}{3^2}\).

That is the reason for multiplying the initial equation \(2^{k-1}*3^{k+1}\) with \(\frac{3^2}{3^2}\)

Hope this helps you!


hey pushpitkc , thank you, yes it helps :-)

can you please explain why do we need to multiply \(2^{k-1}*3^{k+1}\) with \(\frac{3^2}{3^2}\) ? what is the reason ? logic?

thank you and have a great evening :)


Hey dave13

If you look at the answer options, they have either \(3^{k-1}\) or \(6^{k-1}\) in them, but the expression is in the form \(3^{k+1}\).

In order to make the expression resemble the answer option, we use \(\frac{3^2}{3^2}\)

Hope it is clear!
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Re: Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 16 Feb 2019, 22:40
I still don't get the logic of this question, is this a backsolving type of question?

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Re: Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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New post 16 Feb 2019, 23:12
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jfranciscocuencag wrote:
I still don't get the logic of this question, is this a backsolving type of question?

Kind regards!



Hey jfranciscocuencag

No, it is a case of manipulation :dazed

\(2^{k-1}3^{k+1}\)

Now we can write the above as 2^k / 2 * 3 * 3^k

Which becomes 3/2 * \(2^k * 3^k\)

Now if we go over each option, we can try to get 3/2 from one

B actually gives us that

\(3^2\) * \(6^k\) / 6, now \(6^k\)can be written as \(3^k * 2^k\)

3/2 * \(3^k * 2^k\)
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Re: Which of the following is equal to 2^{k-1}3^{k+1}?   [#permalink] 16 Feb 2019, 23:12
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