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Which of the following is equal to 2^{k1}3^{k+1}?
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08 May 2018, 02:52
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[GMAT math practice question] Which of the following is equal to \(2^{k1}3^{k+1}\)? \(A. 2^2(6^{k1})\) \(B. 3^2(6^{k1})\) \(C. 6(2^{k1})\) \(D. 6(3^{k1})\) \(E. 6^{k1}\)
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Re: Which of the following is equal to 2^{k1}3^{k+1}?
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08 May 2018, 03:21
MathRevolution wrote: [GMAT math practice question]
Which of the following is equal to \(2^{k1}3^{k+1}\)?
\(A. 2^2(6^{k1})\) \(B. 3^2(6^{k1})\) \(C. 6(2^{k1})\) \(D. 6(3^{k1})\) \(E. 6^{k1}\) Approach 01 : From Qstem to Answer: \(2^{k1}3^{k+1}\) = \(3^2*(2^{k1}3^{k1})\) = \(3^2(6^{k1})\) .......... Option B is the answerApproach 02 : From Answer to Qstem: Option 01:\(A. 2^2(6^{k1})\) = \(2^2*(2^{k1}*3^{k1})\) = \(2^{k+1}*3^{k1}\) Option 02:\(3^2(6^{k1})\) = \(3^2*(2^{k1}*3^{k1})\) = \(2^{k1}*3^{k+1}\) .......... Option B is the answerOption 03:\(6(2^{k1})\) = \(3*2*(2^{k1})\) = \(3*2^k\) Option 04:\(6(3^{k1})\) = \(3*2*(3^{k1})\) = \(2*3^k\) Option 05:\(6^{k1}\) = \(2^{k1}*3^{k1}\)
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Re: Which of the following is equal to 2^{k1}3^{k+1}?
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08 May 2018, 03:24
MathRevolution wrote: [GMAT math practice question]
Which of the following is equal to \(2^{k1}3^{k+1}\)?
\(A. 2^2(6^{k1})\) \(B. 3^2(6^{k1})\) \(C. 6(2^{k1})\) \(D. 6(3^{k1})\) \(E. 6^{k1}\) \(2^{k1}3^{k+1}\) \(=\) \(2^{k1}*3^{k1}*3^2\) \(=\) \(3^2*6^{k1}\) Answer: B
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Re: Which of the following is equal to 2^{k1}3^{k+1}?
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10 May 2018, 07:58
=> \(2^{k1}3^{k+1} = 2^{k1}3^{k1}*3^2= 3^2*(2*3)^{k1} = 3^2(6^{k1})\) Therefore, the answer is B. Answer: B
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Re: Which of the following is equal to 2^{k1}3^{k+1}?
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10 May 2018, 09:01
MathRevolution wrote: =>
\(2^{k1}3^{k+1} = 2^{k1}3^{k1}*3^2= 3^2*(2*3)^{k1} = 3^2(6^{k1})\)
Therefore, the answer is B. Answer: B how did you solve for 3^k1 * 3^2 ??
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Re: Which of the following is equal to 2^{k1}3^{k+1}?
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23 May 2018, 21:33
SandhyAvinash wrote: MathRevolution wrote: =>
\(2^{k1}3^{k+1} = 2^{k1}3^{k1}*3^2= 3^2*(2*3)^{k1} = 3^2(6^{k1})\)
Therefore, the answer is B. Answer: B how did you solve for 3^k1 * 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 Posted from my mobile device
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Re: Which of the following is equal to 2^{k1}3^{k+1}?
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03 Jun 2018, 05:26
MathRevolution wrote: =>
\(2^{k1}3^{k+1} = 2^{k1}3^{k1}*3^2= 3^2*(2*3)^{k1} = 3^2(6^{k1})\)
Therefore, the answer is B. Answer: B pushpitkc any idea where from do we get additional \(3^2\) here \(2^{k1}3^{k1}* 3^2\) ? thank you anf happy Sunday



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Which of the following is equal to 2^{k1}3^{k+1}?
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03 Jun 2018, 06:25
dave13 wrote: MathRevolution wrote: =>
\(2^{k1}3^{k+1} = 2^{k1}3^{k1}*3^2= 3^2*(2*3)^{k1} = 3^2(6^{k1})\)
Therefore, the answer is B. Answer: B pushpitkc any idea where from do we get additional \(3^2\) here \(2^{k1}3^{k1}* 3^2\) ? thank you anf happy Sunday Hey dave13\(2^{k1}*3^{k+1} = 2^{k1}*3^{k+1}*\frac{3^2}{3^2} = 2^{k1}*3^{k+1}*3^{2}*3^2\) \(3^{k+1}*3^{2}\) is of form \(a^m * a^n\) which is equal to \(a^{m+n}\). Here, we get \(3^{k+12} = 3^{k1}\) This further simplifies as follows: \(2^{k1}3^{k1}*3^2= 3^2*(2*3)^{k1} = 3^2(6^{k1})\) Hope this helps you!
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Re: Which of the following is equal to 2^{k1}3^{k+1}?
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03 Jun 2018, 07:15
pushpitkc wrote: dave13 wrote: MathRevolution wrote: =>
\(2^{k1}3^{k+1} = 2^{k1}3^{k1}*3^2= 3^2*(2*3)^{k1} = 3^2(6^{k1})\)
Therefore, the answer is B. Answer: B pushpitkc any idea where from do we get additional \(3^2\) here \(2^{k1}3^{k1}* 3^2\) ? thank you anf happy Sunday Hey dave13\(2^{k1}*3^{k+1} = 2^{k1}*3^{k+1}*\frac{3^2}{3^2} = 2^{k1}*3^{k+1}*3^{2}*3^2\) \(3^{k+1}*3^{2}\) is of form \(a^m * a^n\) which is equal to \(a^{m+n}\). Here, we get \(3^{k+12} = 3^{k1}\) This further simplifies as follows: \(2^{k1}3^{k1}*3^2= 3^2*(2*3)^{k1} = 3^2(6^{k1})\) 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 \(2^{k1}*3^{k+1}\) there is no such number



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Re: Which of the following is equal to 2^{k1}3^{k+1}?
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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 \(2^{k1}*3^{k+1}\) there is no such number Hey dave13Any 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 \(2^{k1}*3^{k+1}\) with \(\frac{3^2}{3^2}\) Hope this helps you!
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Re: Which of the following is equal to 2^{k1}3^{k+1}?
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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 \(2^{k1}*3^{k+1}\) there is no such number Hey dave13Any 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 \(2^{k1}*3^{k+1}\) 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 \(2^{k1}*3^{k+1}\) 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^{k1}3^{k+1}?
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03 Jun 2018, 11:53
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 \(2^{k1}*3^{k+1}\) there is no such number Hey dave13Any 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 \(2^{k1}*3^{k+1}\) 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 \(2^{k1}*3^{k+1}\) with \(\frac{3^2}{3^2}\) ? what is the reason ? logic? thank you and have a great evening Hey dave13If you look at the answer options, they have either \(3^{k1}\) or \(6^{k1}\) 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^{k1}3^{k+1}?
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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^{k1}3^{k+1}?
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16 Feb 2019, 23:12
jfranciscocuencag wrote: I still don't get the logic of this question, is this a backsolving type of question?
Kind regards! Hey jfranciscocuencagNo, it is a case of manipulation \(2^{k1}3^{k+1}\) 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^{k1}3^{k+1}?
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