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

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

by Brent@GMATPrepNow » Thu Feb 02, 2017 6:48 am
Here's a 700+ level question I just created.
If 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
Answer: D
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by regor60 » Thu Feb 02, 2017 7:08 am
Brent@GMATPrepNow wrote:Here's a 700+ level question I just created.
If 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
Answer: D
Are logarithms allowed ? If so, straightforward

Anyhow, using info provided,

27/16 = 3^3/2^4 = 2^j/3^k = 3^-k/2^-j, therefore

3^3/2^4 = 3^-k/2^-j. Equating the exponents, k=-3 and j=-4, therefore kj=

12

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by Brent@GMATPrepNow » Thu Feb 02, 2017 7:24 am
regor60 wrote:
Are logarithms allowed ? If so, straightforward
There's no way that you can be prevented from using logarithms on test day, but you are not permitted to use a calculator on the quantitative section, so logarithms won't help you much.

Cheers,
Brent
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by Brent@GMATPrepNow » Thu Feb 02, 2017 7:28 am
regor60 wrote:
Brent@GMATPrepNow wrote:Here's a 700+ level question I just created.
If 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
Answer: D
Are logarithms allowed ? If so, straightforward

Anyhow, using info provided,

27/16 = 3^3/2^4 = 2^j/3^k = 3^-k/2^-j, therefore

3^3/2^4 = 3^-k/2^-j. Equating the exponents, k=-3 and j=-4, therefore kj=

12
Your final answer is correct. However, it is not true that k = -3 and j = -4.
If we plug those values into the given information (3^k = 16, and 2^j = 27), we find that they are not solutions.

For example, 3^(-3) = 1/27, not 16
Likewise, 2^(-4) = 1/16, not 27

Cheers,
Brent
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by GMATGuruNY » Thu Feb 02, 2017 8:27 am
Nice problem!
If 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
A number property rule:
If (a^r)(b^s)(c^t) = (a^x)(b^y)(x^z), then rst = xyz.

Multiplying 3^k = 16 by 2^j = 27, we get:
(3^k)(2^j) = (16)(27)
(2^j)(3^k) = 2�3³.

Applying the number property rule above, we get:
jk = 4*3 = 12.

The correct answer is D.
Last edited by GMATGuruNY on Fri Feb 03, 2017 9:19 am, edited 2 times in total.
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by Brent@GMATPrepNow » Thu Feb 02, 2017 9:10 am
Brent@GMATPrepNow wrote: If 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
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

Answer: D

Cheers,
Brent
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by Brent@GMATPrepNow » Thu Feb 02, 2017 1:03 pm
Brent@GMATPrepNow wrote:If 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
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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by Brent@GMATPrepNow » Thu Feb 02, 2017 1:26 pm
regor60 wrote:
Brent@GMATPrepNow wrote:Here's a 700+ level question I just created.
If 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
Answer: D
Are logarithms allowed ? If so, straightforward
Here's a solution that involves logarithms AND doesn't require a calculator.
NOTE: This is wayyyyyy beyond the scope of the GMAT. But, if you happen to know a few logarithm rules, we can solve this question quickly.

Given: 3^k = 16
Take log of both sides: log(3^k) = log16
Apply log of power law: klog3 = log16
Solve for k: k = (log16)/(log3)

Given: 2^j = 27
Take log of both sides: log(2^j) = log27
Apply log of power law: jlog2 = log27
Solve for j: j = (log27)/(log2)

So, jk = (log27/log2) x (log16)/(log3)
= (log27 x log16)/(log2 x log3)
= (log27)/(log3) x (log16)/(log2)
= log327 x log216
= 3 x 4
= 12
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by Matt@VeritasPrep » Thu Feb 02, 2017 5:14 pm
I've got a few! First one is easiest: (crudely) approximate.

3� = 16, so 2 < k < 3, close to 2.5

2ʲ = 27, so 4 < j < 5, close to 4.75 or so

So we're around 2.5 * 4.75, and only D fits.

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by Matt@VeritasPrep » Thu Feb 02, 2017 5:16 pm
Now for a sleazy one that I really like:

Start by dividing each equation to set them each equal to 1:

3�/16 = 1

2ʲ/27 = 1

Since they're each equal to 1, they must be equal to each other:

3�/16 = 2ʲ/27

Cross-multiply:

3� * 27 = 2ʲ * 16

3^(k+3) = 2^(j+4)

If each exponent = 0, we've got one "solution", so (cough cough) "k = -3", "j = -4", and j*k = 12.

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by Matt@VeritasPrep » Thu Feb 02, 2017 5:22 pm
We could also trick out some logarithms here. Let's say that 2Ë£ = 3. Then we've got:

(2ˣ)� = 16, or xk = 4

Let's also say that 3ʸ = 2. Then we've got

(3ʸ)ʲ = 27, or yj = 3

From this, we've got x * k * y * j = 4 * 3. But we also know that x * y = log3/log2 * log2/log3 = 1, so we're left with k * j = 12.

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by Matt@VeritasPrep » Thu Feb 02, 2017 5:29 pm
While we're talking logarithms, here's a problem I really like from an old AHSME (the precursor to the AMC). The problem seeks the value of q/p, and can be solved with surprisingly simple algebra ... but how? :D

log₉p = log�₂q = log�₆(p+q)

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by Matt@VeritasPrep » Thu Feb 02, 2017 5:48 pm
Matt@VeritasPrep wrote:We could also trick out some logarithms here. Let's say that 2Ë£ = 3. Then we've got:

(2ˣ)� = 16, or xk = 4

Let's also say that 3ʸ = 2. Then we've got

(3ʸ)ʲ = 27, or yj = 3

From this, we've got x * k * y * j = 4 * 3. But we also know that x * y = log3/log2 * log2/log3 = 1, so we're left with k * j = 12.
Just realized that I can keep this all in the scope of the GMAT and avoid logarithms altogether.

Since 3ʸ = 2 and 2ˣ = 3, we can replace 2 with 3ʸ in the second equation, giving us

(3ʸ)ˣ = 3, or x*y = 1. Ha! No logs in this cabin!

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by regor60 » Fri Feb 03, 2017 6:26 am
Matt@VeritasPrep wrote:Now for a sleazy one that I really like:

Start by dividing each equation to set them each equal to 1:

3�/16 = 1

2ʲ/27 = 1

Since they're each equal to 1, they must be equal to each other:

3�/16 = 2ʲ/27

Cross-multiply:

3� * 27 = 2ʲ * 16

3^(k+3) = 2^(j+4)

If each exponent = 0, we've got one "solution", so (cough cough) "k = -3", "j = -4", and j*k = 12.

According to Brent, k=-3 and j=-4 aren't valid solutions....;)

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by Brent@GMATPrepNow » Fri Feb 03, 2017 6:46 am
regor60 wrote:
Matt@VeritasPrep wrote:Now for a sleazy one that I really like:

Start by dividing each equation to set them each equal to 1:

3�/16 = 1

2ʲ/27 = 1

Since they're each equal to 1, they must be equal to each other:

3�/16 = 2ʲ/27

Cross-multiply:

3� * 27 = 2ʲ * 16

3^(k+3) = 2^(j+4)

If each exponent = 0, we've got one "solution", so (cough cough) "k = -3", "j = -4", and j*k = 12.

According to Brent, k=-3 and j=-4 aren't valid solutions....;)
I should rephrase that.

k=-3 and j=-4 are valid "solutions" . . . to equations that are different from the original equations (3^k = 16, and 2^j = 27)

Then again, I suppose every number is a solution to some equation (many equations actually) :-)
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