I was flipping through the Gmat Math Bible and got to the exponents sections and can't figure out how to solve these exponent inequalties algabraeically. Can anyone please show me how because this concept shows up on hard data suff problems. THanks.

x^2 < x
x^3 > x
x^3 < x^2

1. x^2<x
1.1 for x>0: x^2<x --> x<1 --> x e (0,1)
1.2 for x<0: x^2<x --> x>1 --> no x
1.3. Therefore, x e (0,1)

2. x^3>x
2.1 for x>0: x^3>x --> x^2>1 --> x e (1,+∞)
2.2 for x<0: x^3>x --> x^2<1 --> x e (-1,0)
2.3. Therefore, x e (-1,0)&(1,+∞)

3. x^3<x^2
3.1 x<1 --> x e (-∞,1)
3.2. Therefore, x e (-∞,1)

walker, i have a question re: 2.2

x^2<1 .... practically, i know that only numbers btwn -1 and 0 will satisfy this, but if i were to just look at the statement, i would simplify it to that x < 1 ... ie. any number for x<1 will work.

whats the logic behind avoiding this trap ?

Pmenon, I saw that many people make mistakes in inequalities. I guess these advices can help:

1. pick very large and vary small numbers. It helps recognize mistakes faster.
2. use drawing. It is really helpful and makes a solution obvious. For example, x^2<1 ==> bottom part of the parabola y=x^2 limited by the horizontal line: y=1

yeah but verbally what is the rule cause I cannot see it in your illustration. It seems confusing and I've done these problems a millions different ways but just cant spot the rules.

I get confused because I'm not sure when its the upper limit or the lower limit or if its both. For example :
x^2<x
x<0 x<1
X>0 X>1
Answer: 0<x<1 Why though?

x^2>x
x>0 x>1
x<0 x<1
Answer: x<0 and x>1 why? If I knew the rule I could figure it out. I try all combo's but nothing seems to work.

The main rule is:

if we divide left and right sides of an inequality by a positive number, the sign of the inequality remains the same one.
if we divide left and right sides of an inequality by a negative number, the sign of the inequality changes.

Therefore, when we divide an inequality by x than can be both positive and negative, we have to consider two possibilities and also be careful with x=0.

Consider an example:

x^2>x

for x>0: x^2/x > x/x --> x>1. Therefore, x e (1,+∞)
for x<0: x^2/x < x/x --> x<1. Therefore, x e (-∞,0)
for x=0: x does not satisfy the inequality.

Join three findings: x e (-∞,0)&(1,+∞)

Consider other example:

x^2<x

for x>0: x^2/x < x/x --> x<1. Therefore, x e (0,1)
for x<0: x^2/x > x/x --> x>1. There is no x that satisfies the inequality.
for x=0: x does not satisfy the inequality.

Join three findings: x e (0,1)

I'm missing something here...I don't quite follow '3' above. Can someone elaborate further?

x^3 < x^2
Dividing both sides by x^2,
we get x<1 which means x takes any value from -infinity upto 1.

x^3>x.. i think the range of this is x>-1

x^3<x^2 i think x<-1 but u have a -1<x<0 where this doesnt..

for example, x=0.5 and 0.125>0.5 is not true


for x<0 x^3 is negative and x^2 is positive, so x^3<x^2 is true for x<0
but I found my mistake here: should be x e (-∞,0)&(0,1) rather than x e (-∞,1)