7. Which of these has the smallest value?
A $2016^{-1}$
B $2016^{-1 / 2}$
C $2016^{0}$
D $2016^{1 / 2}$
E $2016^{1}$
Solution
A
Method 1
We have $2016^{-1}=\frac{1}{2016}$ and $2016^{-1 / 2}=\frac{1}{\sqrt{2016}}$. Now, as $1<\sqrt{2016}<2016$, it follows that $\frac{1}{2016}<\frac{1}{\sqrt{2016}}<1$, and hence $2016^{-1}<2016^{-1 / 2}<1$.
On the other hand, $2016^{0}$ is equal to 1 , and both $2016^{1 / 2}=\sqrt{2016}$ and $2016^{1}=2016$ are greater than 1 .
It follows that, of the given options, $2016^{-1}$ has the smallest value.

Method 2
The function $2016^{x}$ may be defined by the equation
$$
2016^{x}=e^{x \ln 2016},
$$
where $e^{x}$ is the exponential function [see the note below] and $\ln 2016$ is the natural logarithm of 2016. (The value of $\ln 2016$ is 7.6089 to 4 decimal places.)

The exponential function is an increasing function in the sense that if $x<x^{\prime}$, then $e^{x}<e^{x^{\prime}}$. It follows that $2016^{x}$ is also an increasing function, because if $x<x^{\prime}$, then $x \ln 2016<x^{\prime} \ln 2016$ and hence $e^{x \ln 2016}<e^{x^{\prime} \ln 2016}$.
Therefore, as $-1<-\frac{1}{2}<0<\frac{1}{2}<1$, we have $2016^{-1}<2016^{-1 / 2}<2016^{0}<2016^{1 / 2}<2016^{1}$. Hence, of the given options, $2016^{-1}$ is the smallest.