Problem with exponents and inequalities

Question

Hello I tried to solve this problem, below:

Given the three numbers $x,y=x^x,z=x^{x^x}$ with $.9<x<1.0$. Arranged in order of increasing magnitude, they are:

$\text{(A) } x,z,y\quad \text{(B) } x,y,z\quad \text{(C) } y,x,z\quad > \text{(D) } y,z,x\quad \text{(E) } z,x,y$

I know the answer is A. I found the solution which is :

since $0<x<1$ then ${x^x} > x$ and since $x > 0$ and $x < 1$ then ${x^x} < 1 $. Thus ${x^{(x^x)}} > x$, so $z > x$. Since $x < {x^x}$ and $x < 1$, we find ${x^x} > {x^{(x^x)}}$, so $y > z$ hence, in increasing order the quantities are $x,z,y$

I'm so confused with this step if someone here could explain in details:

Why or how did we jump from: [ since $x < {x^x}$ and $x < 1$ ] to say that [we find ${x^x} > {x^{(x^x)}}$]

Answer 1

Define a function $f:\mathbb{R^+}\to\mathbb{R^+}$ by $f(t)=x^t$. Since $0<x<1$, we have $$\forall t_1>t_2\in\mathbb{R^+}, f(t_1)=x^{t_1}=x^{t_2}x^{t_1-t_2}<x^{t_2}=f(t_2)$$

This shows that $f(t)$ is strictly decreasing. Hence, since we know that $x<x^x$, we know that $f(x)>f(x^x)$, giving $x^x>x^{x^x}$.

Answer 2

The exponential function $a^x$ with a base $a<1$ is decreasing. Hence if $a<a^a \implies a^a>a^{(a^a)}$