Bunuel wrote:

Is x^5 > x^4?

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.

Hi Bunuel,

From 2 it seems we are able to derive that x cannot equal 0 since 1/x < x and 1< x^2. Why then do we not just do as follows:

1 < x^2

0 < x^2 - 1

(x-1)(x+1) > 0

So, x>1 or x<-1.

Also, keeping in mind that:

1/x < x , then x>0.

Then, from the above x>1.

My tangential questions here would be how did you know to multiply both sides by x^2?

Thanks in advance.

Yes, since x is in the denominator, then it cannot be 0 but we still don't know its sign.

If x > 0, then when cross-multiplying 1/x < x we'd get 1 < x^2.

If x < 0, then when cross-multiplying 1/x < x we'd get 1 > x^2 (flip the sign when multiplying by negative value).

Next, when we multiply an inequality by a variable in even power (x^2, x^4, ...), we don't have to worry about its sing because a number in even power cannot be negative.