Answer: DIf 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
If 3^k = 16, and 2^j = 27
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Here's a 700+ level question I just created.
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Are logarithms allowed ? If so, straightforwardBrent@GMATPrepNow wrote:Here's a 700+ level question I just created.
Answer: DIf 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
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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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.regor60 wrote:
Are logarithms allowed ? If so, straightforward
Cheers,
Brent
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Your final answer is correct. However, it is not true that k = -3 and j = -4.regor60 wrote:Are logarithms allowed ? If so, straightforwardBrent@GMATPrepNow wrote:Here's a 700+ level question I just created.
Answer: DIf 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
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
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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Nice problem!
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.
A number property rule:If 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
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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Another approach is to isolate the 3 in both equations. Here's what I mean:Brent@GMATPrepNow wrote: If 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
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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Another approach involves approximation.Brent@GMATPrepNow wrote:If 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
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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Here's a solution that involves logarithms AND doesn't require a calculator.regor60 wrote:Are logarithms allowed ? If so, straightforwardBrent@GMATPrepNow wrote:Here's a 700+ level question I just created.
Answer: DIf 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
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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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.
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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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.
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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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.
(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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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?
log₉p = log�₂q = log�₆(p+q)
log₉p = log�₂q = log�₆(p+q)
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Just realized that I can keep this all in the scope of the GMAT and avoid logarithms altogether.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.
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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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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I should rephrase that.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....
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)