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# If 3^k = 16, and 2^j = 27, then kj =

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Joined: 12 Sep 2015
Posts: 2703
If 3^k = 16, and 2^j = 27, then kj =  [#permalink]

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02 Feb 2017, 07:50
Top Contributor
7
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15% (low)

Question Stats:

79% (01:08) correct 21% (01:48) wrong based on 161 sessions

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If $$3^k$$ = 16, and $$2^j$$ = 27, then kj =

A) 8
B) 9
C) 10
D) 12
D) 15

* Kudos for all correct solutions

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Brent Hanneson – Founder of gmatprepnow.com

Director
Joined: 05 Mar 2015
Posts: 986
Re: If 3^k = 16, and 2^j = 27, then kj =  [#permalink]

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02 Feb 2017, 10:01
GMATPrepNow wrote:
If $$3^k$$ = 16, and $$2^j$$ = 27, then kj =

A) 8
B) 9
C) 10
D) 12
D) 15

* Kudos for all correct solutions

GMATPrepNow

question must be 3^k=27 && 2^j=16

Any way
3^k=27=3^3 thus k=3
2^j=16=2^4 thus j=4
so kj= 3*4 =12

Ans D
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Joined: 12 Sep 2015
Posts: 2703
Re: If 3^k = 16, and 2^j = 27, then kj =  [#permalink]

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02 Feb 2017, 10:16
1
Top Contributor
rohit8865 wrote:
GMATPrepNow wrote:
If $$3^k$$ = 16, and $$2^j$$ = 27, then kj =

A) 8
B) 9
C) 10
D) 12
D) 15

* Kudos for all correct solutions

question must be 3^k=27 & 2^j=16

The given information is correct as worded: $$3^k$$ = 16, and $$2^j$$ = 27
The solution to each individual equation is not an integer.

Cheers,
Brent
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Brent Hanneson – Founder of gmatprepnow.com

Director
Joined: 05 Mar 2015
Posts: 986
Re: If 3^k = 16, and 2^j = 27, then kj =  [#permalink]

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02 Feb 2017, 10:33
GMATPrepNow wrote:
rohit8865 wrote:
GMATPrepNow wrote:
If $$3^k$$ = 16, and $$2^j$$ = 27, then kj =

A) 8
B) 9
C) 10
D) 12
D) 15

* Kudos for all correct solutions

question must be 3^k=27 & 2^j=16

The given information is correct as worded: $$3^k$$ = 16, and $$2^j$$ = 27
The solution to each individual equation is not an integer.

Cheers,
Brent

then we can multiply both equation to get
3^k*2^j= 16*27
getting k=3 j=4
thus KJ=12
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Re: If 3^k = 16, and 2^j = 27, then kj =  [#permalink]

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02 Feb 2017, 11:22
rohit8865 wrote:
then we can multiply both equation to get
3^k*2^j= 16*27
getting k=3 j=4
thus KJ=12

The question looks a bit odd , but I am with the above solution, answer must be (D)...

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CEO
Joined: 12 Sep 2015
Posts: 2703
Re: If 3^k = 16, and 2^j = 27, then kj =  [#permalink]

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02 Feb 2017, 14:04
1
Top Contributor
3
GMATPrepNow wrote:
If $$3^k$$ = 16, and $$2^j$$ = 27, then kj =

A) 8
B) 9
C) 10
D) 12
D) 15

* Kudos for all correct solutions

Another approach is to isolate the 3 in both equations. Here’s what I mean:

Given: 3^k = 16
Rewrite 16 as 2^4 to get: 3^k = 2^4
Raise both sides to the power of 1/k to get: (3^k)^(1/k) = (2^4)^(1/k)
Use power of power law to simplify: 3 = 2^(4/k)

Given: 2^j = 27
Rewrite 27 as 3^3 to get: 2^j = 3^3
Raise both sides to the power of 1/3 to get: (2^j)^(1/3) = (3^3) ^(1/3)
Use power of power law to simplify: 2^(j/3) = 3

We now have two equations:
3 = 2^(4/k)
2^(j/3) = 3

Since both equations are set equal to 3, we can write: 2^(4/k) = 2^(j/3)
Since the bases both equal 2, we can conclude that 4/k = j/3
Cross multiply to get: jk = (4)(3)
So, jk = 12

Cheers,
Brent
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Re: If 3^k = 16, and 2^j = 27, then kj =  [#permalink]

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02 Feb 2017, 14:05
Top Contributor
2
GMATPrepNow wrote:
If $$3^k$$ = 16, and $$2^j$$ = 27, then kj =

A) 8
B) 9
C) 10
D) 12
D) 15

* Kudos for all correct solutions

Another approach involves approximation.

Given: 2^j = 27
Notice that 2^4 = 16 and 2^5 = 32
Since 27 is closer to 32 than it is to 16, we can conclude that j is closer to 5 than it is to 4.
Let's say j ≈ 4.7

Given: 3^k = 16
Notice that 3^2 = 9 and 3^3 = 27
Since 16 is approximately halfway between 9 and 27, we can conclude that k is approximately halfway between 2 and 3.
Let's say k ≈ 2.5

So, jk ≈ (4.7)(2.5)
≈ 11.75

When we check the answer choices, we see that answer choice D is the closest to 11.75

Cheers,
Brent
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Re: If 3^k = 16, and 2^j = 27, then kj =  [#permalink]

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15 Feb 2017, 10:22
1
GMATPrepNow wrote:
If $$3^k$$ = 16, and $$2^j$$ = 27, then kj =

A) 8
B) 9
C) 10
D) 12
D) 15

* Kudos for all correct solutions

Another approach. 3^k = 16. Take log to the base 3. k= log316 = 4 log32 As loga(b^k) = k logab
2^j = 27. Take log to the base 2. j = log227 = 3log33
k*j = 12 * log32 * log 23
Thus kj = 12. as log ab * logba =1
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Re: If 3^k = 16, and 2^j = 27, then kj =  [#permalink]

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09 Jul 2017, 08:13
1
3^k=16 and 2^J=27, we need to find kJ
3^k x 2^J=16 x 27
3^k x 2^J=2^4 x 3^3
i.e. 3^k x 2^J=3^3 x 2^4
Thus, kJ=12

Ans : D
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Joined: 01 May 2017
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Re: If 3^k = 16, and 2^j = 27, then kj =  [#permalink]

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09 Jul 2017, 09:13
The answer would be D i.e. 12.

Multiply the two and get a he respective powers of 2 and 3.

Thus k*j=12

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Re: If 3^k = 16, and 2^j = 27, then kj =  [#permalink]

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14 Jul 2018, 08:38
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