Is x > k?
(1) 2^x • 2^k = 4
(2) 9^x • 3^k = 81
exponents.
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 66
- Joined: Wed Feb 20, 2008 6:23 pm
- Thanked: 2 times
-
- Master | Next Rank: 500 Posts
- Posts: 259
- Joined: Thu Jan 18, 2007 8:30 pm
- Thanked: 16 times
I am getting "C".
Stmt1;
2^x • 2^k = 4 ==> 2^x • 2^k = 2^2
==> x+ k = 2 insufficient
Stmt 2;
9^x • 3^k = 81 ==> 3^2x.3^k = 3^4
2x+k = 4 --> insufficient
from 1 and 2 ==> x = 2 and k = 0 hence sufficient.
Stmt1;
2^x • 2^k = 4 ==> 2^x • 2^k = 2^2
==> x+ k = 2 insufficient
Stmt 2;
9^x • 3^k = 81 ==> 3^2x.3^k = 3^4
2x+k = 4 --> insufficient
from 1 and 2 ==> x = 2 and k = 0 hence sufficient.
I am getting C
Maybe I am missing something
Assumption . means multiplication!!!
st1: equate both the sides to the base of 2, the powers should be equal
so, x + k = 2, no info on x and k, insufficient
st2: equate both the sides to the base of 3, the powers should be equal
so, 2x + k = 4, no info on x and k, insufficient
Together: 2 variables, 2 equations: SUFFICIENT
Maybe I am missing something
Assumption . means multiplication!!!
st1: equate both the sides to the base of 2, the powers should be equal
so, x + k = 2, no info on x and k, insufficient
st2: equate both the sides to the base of 3, the powers should be equal
so, 2x + k = 4, no info on x and k, insufficient
Together: 2 variables, 2 equations: SUFFICIENT
-
- Legendary Member
- Posts: 833
- Joined: Mon Aug 04, 2008 1:56 am
- Thanked: 13 times