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

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

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08 May 2018, 02:52
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Question Stats:

76% (01:40) correct 24% (02:07) wrong based on 76 sessions

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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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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" RC Moderator Joined: 24 Aug 2016 Posts: 789 GMAT 1: 540 Q49 V16 GMAT 2: 680 Q49 V33 Re: Which of the following is equal to 2^{k-1}3^{k+1}? [#permalink] ### Show Tags 08 May 2018, 03:21 1 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:$$6^{k-1}$$ = $$2^{k-1}*3^{k-1}$$ _________________ Please let me know if I am going in wrong direction. Thanks in appreciation. Manager Joined: 28 Nov 2017 Posts: 141 Location: Uzbekistan Re: Which of the following is equal to 2^{k-1}3^{k+1}? [#permalink] ### Show Tags 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 _________________ Kindest Regards! Tulkin. Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8017 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Which of the following is equal to 2^{k-1}3^{k+1}? [#permalink] ### Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: Which of the following is equal to 2^{k-1}3^{k+1}?  [#permalink]

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

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

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

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

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

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

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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03 Jun 2018, 06:25
1
1
dave13 wrote:
MathRevolution wrote:
=>

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

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

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

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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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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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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03 Jun 2018, 11:53
1
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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16 Feb 2019, 22:40
I still don't get the logic of this question, is this a backsolving type of question?

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

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16 Feb 2019, 23:12
1
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

$$2^{k-1}3^{k+1}$$

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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