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

Dear Experts,
I wonder how Mitch reaches that j=4 & k=3 and does not satisfies individual equations, while in Brent's first solution proves that J & K are not integer individually because 3 = 2^(4/k) & 2^(j/3) = 3.

Can any expert elaborate more as question seems strange?

thanks

Mo2men wrote:Dear Experts,
I wonder how Mitch reaches that j=4 & k=3 and does not satisfies individual equations, while in Brent's first solution proves that J & K are not integer individually because 3 = 2^(4/k) & 2^(j/3) = 3.

Can any expert elaborate more as question seems strange?

thanks

In my solution above, I had in mind the following number property rule:
If (a^r)(b^s)(c^t) = (a^x)(b^y)(c^z), then rst = xyz.
I've revised my post to clarify the reasoning.

A logarithmic proof for this number property rule:

(a^r)(b^s)(c^t) = (a^x)(b^y)(c^z)

log(a^r) * log(b^s) * log(c^t) = log(a^x) * log(b^y) * log(c^z)

r(log a) * s(log b) * t(log c) = x(log a) * y(log b) * z(log c)

rst = xyz.

Please note that this proof is beyond the scope of the GMAT.

Wonderful explanations from big experts.

thanks for Brent and Mitch

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

3^k = 16
Taking Log on both sides,
-> k log3 = log 16 = 4 log2
-> k = 4 log2 /log3 ..................(i)

similarly , 2^j = 27
-> j log2 = log 27 = 3 log3
-> j = 3 log3/log2 .....................(ii)

Multiplying (i) & (ii) we get
kj = 4 x 3 = 12

Hence Option D

Mo2men wrote:Dear Experts,
I wonder how Mitch reaches that j=4 & k=3 and does not satisfies individual equations, while in Brent's first solution proves that J & K are not integer individually because 3 = 2^(4/k) & 2^(j/3) = 3.

Can any expert elaborate more as question seems strange?

thanks

This is all beyond the GMAT, so anyone not interested in it can safely ignore this post.

Part of the issue here is that 3� = 16 has an infinite number of complex solutions. Any number of the form (4*ln(2) + 2*i*π*n) / ln(3), where n is an integer, is a solution to the equation.

Leaving that aside, though, in my second solution below, the bug is that 3�/16 = 2ʲ/27 only transforms to 3^(k+3) = 2^(j+4) if k and j are both POSITIVE. But that transformation still allows us to get the product of j and k! So it's a sneaky technicality to get what the problem seeks - (j * k) - without providing valid individual j and k.

Expounding on that a bit further, if k and j are not positive (but simply real numbers), 3�/16 = 2ʲ/27 transforms to

k = (j * ln(2) + 4 * ln(2) - 3 * ln(3)) / ln(3)

From here, j * k is just

j * (j * ln(2) + 4 * ln(2) - 3 * ln(3)) / ln(3) =>

Since we know 2ʲ = 27, we can replace j above with 3 * ln(3) / ln(2), and after a lot of boring substitution, find that j * k = 12.