### Operations for functions: Exponential and logarithmic functions

### Growth of a logarithmic function

Below, the graphs of the logarithmic functions #\log_2# and #\log_{\frac{1}{2}}# are drawn in blue.

As the logarithm is the inverse of the exponential, obtaining its graph is simple; all you need to do is mirror the graph of the exponential function (drawn in red) around the line \(y=x\).

These graphs show that #\log# is monotonic and that it grows less fast than a polynomial.

Logarithmic versus polynomial growth rule

For \(a\gt1\) and \(n\in\mathbb{N}\) \[\lim_{x\to\infty}\frac{x^n}{\log_a(x)}=\infty\] and \[\lim_{x\to\infty}\frac{\log_a(x)}{x^n}=0\]

In other words, the term \(x^n\) becomes much bigger than \(\log_a(x)\) (for any given natural number \(n\)), when \(x\) tends to infinity (the term \(x^n\) "grows faster" than the term \(\log_a(x)\)). Notice that this holds regardless how close #a# is to #1# and regardless how large #n# is.

This result is just a reformulation of the theorem *Exponential versus polynomial growth*, as can be seen by substituting #x=a^y# in the above expressions. For instance, the first limit then becomes #\lim_{y\to\infty}\frac{a^{y\cdot n}}{\log_a(a^y)}=\lim_{y\to\infty}\frac{\left(a^y\right)^ n}{y}=\left(\lim_{y\to\infty}\frac{a^{y}}{y^{\frac{1}{n}}}\right)^n=\infty#.

This follows from rule 6 of

*Rules for Limits*with #\displaystyle f(x)=\dfrac{x^{11}}{2^{8 x}}# and #h(x) = \log_2(x)#. By applying the theory to #\displaystyle \lim_{x\to\infty}f(x) # we find that the limit is #0#.

**Pass Your Math**independent of your university. See pricing and more.

Or visit omptest.org if jou are taking an OMPT exam.