Question

Let a,b,x be three real numbers with a>b and x>0. Which of the following statements is correct?

A. X^a>X^b if a,b>1 and for every x>0

B. X^a<X^b if X is an element of (0,1)

C.X^a<X^b if a, b>0 and for every x>1

D.X^a>X^b if a,b x>0

Answer 1

A.

$$x^a>x^b, a>b>1,x>0$$

 False.

Counterexample

$$x=\dfrac{1}{2}>0, a=3>1, b=2>1, 3>2$$

$$x^a=(\dfrac{1}{2})^3=\dfrac{1}{8}, x^b=(\dfrac{1}{2})^2=\dfrac{1}{4}$$

$$\dfrac{1}{8}<\dfrac{1}{4}=>x^a<x^b$$

B.

$$x^a<x^b, a>b,0<x<1$$

 True.

The function $f(t)=x^t$ is decreasing, if $0<x<1.$ Then

$$x^a<x^b, \text{if }a>b$$

C.

$$x^a<x^b, a>b>0,x>1$$

 False.

Counterexample

$$x=2>1, a=3>0, b=2>0, 2>1$$

$$x^a=(2)^3=8, x^b=(2)^2=4$$ $$8>4=>x^a>x^b$$

D.

$$x^a>x^b, a>b>0,x>0$$

 False.

Counterexample

$$x=\dfrac{1}{2}>0, a=3>0, b=2>0, 3>2$$

$$x^a=(\dfrac{1}{2})^3=\dfrac{1}{8}, x^b=(\dfrac{1}{2})^2=\dfrac{1}{4}$$

$$\dfrac{1}{8}<\dfrac{1}{4}=>x^a<x^b$$