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

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Re: If 3^k = 16, and 2^j = 27, then kj = [#permalink]
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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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Re: If 3^k = 16, and 2^j = 27, then kj = [#permalink]
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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]
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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Re: If 3^k = 16, and 2^j = 27, then kj = [#permalink]
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Top Contributor
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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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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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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Re: If 3^k = 16, and 2^j = 27, then kj = [#permalink]
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]
Top Contributor
Given that $$3^k$$ = 16, and $$2^j$$ = 27 and we need to find the value of kj

$$3^k$$ = 16 = $$2^4$$

Raising both the sides to the power of $$\frac{1}{k}$$ we get
$$3^{k/k}$$ = $$2^{\frac{4}{k}}$$
=> 3 = $$2^{\frac{4}{k}}$$ ...(1)

$$2^j$$ = 27 = $$3^3$$
Raising both the sides to the power of $$\frac{1}{3}$$ we get
$$2^{\frac{j}{3}}$$ = $$3^{\frac{3}{3}}$$
=> $$2^{\frac{j}{3}}$$ = 3 ...(2)

From (1) and (2) we get
3 = $$2^{\frac{4}{k}}$$ = $$2^{\frac{j}{3}}$$
=> $$\frac{4}{k}$$ = $$\frac{j}{3}$$
=> 4 * 3 = k*j
=> kj = 12

Hope it helps!

Watch the following video to learn the Basics of Exponents

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