If \(a > b\), is it always true that \(a^n > b^n?\)
True or False?
If \(a>b\), then \(a^n>b^n.\)
Why some people say it's true: The more you multiply, the bigger it gets, so obviously if \(a>b\), \(a^n>b^n\).
Why some people say it's false: It doesn't work for all numbers.
This statement is \( \color{red} {\textbf{false}}\). In particular, if \(a>b>0\), then \[\begin{cases} a^n>b^n,~ \text{if } n>0; \\ a^n<b^n,~ \text{if } n<0; \\ a^n=b^n,~ \text{if } n=0.\end{cases}\]
Proof:
First, we need to understand the graph of the function \(f(x)=k^x (k\neq0)\).
As shown in the figure above, we have 3 cases:
If \(k>1\), the graph of \(f\) is monotonically increasing; \(f(x)\geqslant1\) when \(x\geqslant0\), and \(f(x)<1\) when \(x<0\).
If \(k<1\), the graph of \(f\) is monotonically decreasing; \(f(x)\leqslant1\) when \(x\geqslant0\), and \(f(x)>1\) when \(x<0\).
If \(k=1\), then \(f(x)=1\).Now consider 3 functions \(f(x)=a^x\), \(g(x)=b^x\) and \(h(x)=\frac{f(x)}{g(x)},\) where \(a>b>0.\) Then we have \[h(x)=\frac{a^x}{b^x}=\left(\frac{a}{b}\right)^x.\] As \(a>b>0\) and \(b\neq0\), we have \(\frac{a}{b}>1\).
Hence, \(h(x)=\left(\frac{a}{b}\right)^x\) is in the form \(k^x\) and \(k=\frac{a}{b}>1\), so the graph of \(h(x)\) is monotonically increasing.When \(x>0\), we have \(h(x)=\frac{f(x)}{g(x)}>1\). Since \(g(x)>0\), \[f(x)>g(x) \implies a^x>b^x.\]
When \(x<0\), we have \(h(x)=\frac{f(x)}{g(x)}<1\). Since \(g(x)>0\), \[f(x)<g(x) \implies a^x<b^x.\]
When \(x=0\), we have \(h(x)=\frac{f(x)}{g(x)}=1\). Thus, \[f(x)=g(x) \implies a^x=b^x. \ _\square\]
Examples
Let \(a=3, b=2,\) and \(n=-1\). Find the relationship between \(a^n\) and \(b^n\).
We have \[\begin{align} a^n &= 3^{-1}= 0.33\ldots\\ b^n &= 2^{-1} = 0.5. \end{align}\] So, \(a^n < b^n\). \(_\square\)
Let \(a=3, b=2,\) and \(n=0\). Find the relationship between \(a^n\) and \(b^n\).
We have \[\begin{align} a^n &= 3^0 = 1\\ b^n &= 2^0 = 1. \end{align}\] So, \(a^n = b^n\). \(_\square\)
So, we see that in spite of \(a\) being greater than \(b\), \(a^n\) is not always greater than \(b^n\).