Explain please

gmater29 wrote:Is x > k?
(1) 2^x • 2^k = 4
(2) 9^x • 3^k = 81

IMO, it should be E.

1) 2^x * 2^k = 4
2^(x+k) = 2^2
x+k = 2

Insufficient.

2) 9^x * 3^k = 81
3^2x * 3^k = 3^4
3^(2x+k) = 3^4
2x+k = 4

Insufficient.

Combining, solve the 2 eqns simultaneously
But you get two k values here. x=2,
and k=2 if you sub in 1st eqn, and k=0 if you sub in 2nd eqn.

Insufficient.

Hence E. Correct me if i'm wrong in any step

I think it is C instead because you have 2 equations so you can set up a system of 2 equations with 2 unknown vars.

And it gives you x=2 and k = 0

Do you agree ?

I think it's C

A) 2^x * 2^k = 4
Combine both statements: 2^(x+k) = 4. X can be 0, K can be 2 or X can be 2 or K can be zero. Not sufficient

2) 9^x * 3^k = 81. Simplify: 3^2x * 3^k = 81 so 3^(2x+k) = 81. So 2x+k = 4. X can be 2, K can be 0 or X can be 1, K can be 2 or X can be 0, K can be 4. Not sufficient.


If we combine and solve for two equations: x+k = 2 and 2x+k = 4 then X = 0 and K = 2. Sufficient.

What's the OA?

OA C

Thanks all

adamsmith2009 wrote:I think it's C

A) 2^x * 2^k = 4
Combine both statements: 2^(x+k) = 4. X can be 0, K can be 2 or X can be 2 or K can be zero. Not sufficient

2) 9^x * 3^k = 81. Simplify: 3^2x * 3^k = 81 so 3^(2x+k) = 81. So 2x+k = 4. X can be 2, K can be 0 or X can be 1, K can be 2 or X can be 0, K can be 4. Not sufficient.


If we combine and solve for two equations: x+k = 2 and 2x+k = 4 then X = 0 and K = 2. Sufficient.

What's the OA?

If we multiply x + k = 2 by -1 and subtract it from 2x + k = 4 shouldn't x be 2 and k =0 after

You are right. It should be x=2 and k=0. But anyway it doesn't matter for data sufficiency. Had it been a problem solving question, then it matters

papgust wrote:You are right. It should be x=2 and k=0. But anyway it doesn't matter for data sufficiency. Had it been a problem solving question, then it matters

I agree that it doesn't matter here. Only wanted to make sure i was solving simultaneous equations correctly. Thanks for responding.

Yes - my mistake - X = 2 K = 0