DS - Veritas Prep Algebra Question 62

[scoowhoop]

Is 2^x greater than 100?

(1) 2^(sqrt(x)) = 8
(2) 1/2^x < 0.01

[spoiler](1) 2^(sqrt(x)) = 2^3 --> sqrt(x) = 3 --> x = 9 --> 2^9 is greater than 100; SUFFICIENT
(2) 1/2^x < 0.01 --> 2^x > 1/0.01 --> 2^x > 100; SUFFICIENT

ANS: D

I understand the method in (2) but why the sign change? I thought this only happened when you multiply or divide by a negative number or a varible that is negative? Please help explain! And yes I know it is not relevant to answer the question since you only need to know if you can solve for x. This is just for greater understanding of inequalities. Thanks![/spoiler]

[ajith]

Is 2^x greater than 100?

(1) 2^(sqrt(x)) = 8
(2) 1/2^x < 0.01

[spoiler](1) 2^(sqrt(x)) = 2^3 --> sqrt(x) = 3 --> x = 9 --> 2^9 is greater than 100; SUFFICIENT
(2) 1/2^x < 0.01 --> 2^x > 1/0.01 --> 2^x > 100; SUFFICIENT

ANS: D

I understand the method in (2) but why the sign change? I thought this only happened when you multiply or divide by a negative number or a varible that is negative? Please help explain! And yes I know it is not relevant to answer the question since you only need to know if you can solve for x. This is just for greater understanding of inequalities. Thanks![/spoiler]

The sign change happens when you take reciprocals on both sides [Provided signs are same on both sides]
In this case

1/2^x is positive
0.01 is also positive

so if 1/2^x < 1/100
2^x > 100

[Please be sure that signs are the same on both sides]

[scoowhoop]

Thanks!

[pops]

Is 2^x greater than 100?

(1) 2^(sqrt(x)) = 8
(2) 1/2^x < 0.01

[spoiler](1) 2^(sqrt(x)) = 2^3 --> sqrt(x) = 3 --> x = 9 --> 2^9 is greater than 100; SUFFICIENT
(2) 1/2^x < 0.01 --> 2^x > 1/0.01 --> 2^x > 100; SUFFICIENT

ANS: D

I understand the method in (2) but why the sign change? I thought this only happened when you multiply or divide by a negative number or a varible that is negative? Please help explain! And yes I know it is not relevant to answer the question since you only need to know if you can solve for x. This is just for greater understanding of inequalities. Thanks![/spoiler]

See,
(1/2)^x < 1/100
now, when you take reciprocal you change sign
for example.. 1/2 > 1/3 but 2 < 3

hence, 2^x > 100
Hope, its clear..