1968 AHSME Problems/Problem 29

Problem

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:

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

Solution 1

Seeing that we need to compare values with exponents, we think logarithms. Taking the logarithm base $x$ of each term, we obtain $1$, $x$, and $x^x$. Because $0<x<1$, $f(n)=\log_x(n)$ is monotonically decreasing, so the order of terms by magnitude in our new set of numbers will be reversed compared to the original set (i.e. if $a<b<c$, then $\log_x(a)>\log_x(b)>\log_x(c))$. However, the order of this set will be reversed again (back to the order of the original set) when we take the logarithm base $x$ a second time. After doing this operation, we find the values $0$, $1$, and $x$, which correspond to $x$, $y$, and $z$, respectively. Because $0.9<x<1$, $0<x<1$, and so, by the correspondence detailed above, $x<z<y$, which yields us answer choice $\fbox{A}$.


Solution 2

Because $0<x<1$, taking $x$ to the $x$th power will bring it closer to $1$, thereby raising its value. Because we have established that $x<x^x$, and $f(n)=x^n$ is a monotonically decreasing function, we know that $x^x>x^{x^x}$. However, because $x^x<1$, $x^{x^x}$, compared to $x$, will be closer to (but still less than) $1$. Thus, $x^{x^x}>x$. Putting this all together, we see that $x<x^{x^x}<x^x$, or $x<z<y$, which is answer choice $\fbox{A}$.


See also

1968 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 30
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