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

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

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

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

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

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)