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Answer: DIf 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
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
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
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
A number property rule: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: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.Brent@GMATPrepNow wrote:If 3^k = 16, and 2^j = 27, then kj =
A) 8
B) 9
C) 10
D) 12
D) 15
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
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.
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.
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....
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