Exponentials and logarithms

We continue this topic after the intermediate value theorem posted earlier.

For a>1, define f: \Re \rightarrow \Re by f(x)=a^{x}. It is easily seen that f(x)<f(y) if x>y. This shows that f is one to one. Further, \lim_{x \rightarrow \infty}f(x)=\infty, whereas \lim_{x \rightarrow -\infty}f(x)=0. That, f: \Re \rightarrow \Re is onto \Re_{+} follows from the intermediate value theorem. Thus, f:\Re \rightarrow \Re_{+} defined by f(x)=a^{x} is bijective. So there is a unique map

g: \Re_{+} \rightarrow \Re

such that f(g(y))=y for every y in \Re_{+} and g(f(x))=x for every x in \Re.

This function g is what we call the logarithm function of y to the base a, written as \log_{a}{y}. In fact, the logarithm is a continuous function.

For y_{0} \in \Re_{+}, \varepsilon>0, let \delta=min\{ a^{x_{0}+\varepsilon}-a^{x_{0}}, a^{x_{0}}-a^{x_{0}-\varepsilon}\}, where x_{0}=\log_{a}{y_{0}}. Then, we have for

|y_{0}-y|<b, a^{x_{0}-\varepsilon} \leq y_{0}-\delta <y < y_{0}+\delta \leq a^{x_{0}+\varepsilon}, or

x_{0}-\varepsilon<\log_{a}{y}<x_{0}+\varepsilon or

g(y_{0})-\varepsilon<g(y)<g(y)+\varepsilon

Exercise.

If f:\Re \rightarrow \Re is an increasing continuous function, show that it is bijective onto its range and its inverse is also continuous.

With the help of the logarithm function, we can evaluate \lim_{x \rightarrow 0}\frac{a^{x}-1}{x}.

Let a^{x}=1+y so that y \rightarrow 0 as x \rightarrow 0. Also, x=\log_{a}{1+y}. So, we have

\lim_{x \rightarrow 0}\frac{a^{x}-1}{x}=\lim_{y \rightarrow 0}\frac{y}{\log_{a}{1+y}}=\lim_{y \rightarrow 0}\frac{1}{\frac{1}{y}\log_{a}{1+y}}=\lim_{y \rightarrow 0}\frac{1}{\log_{a}{(1+y)^{\frac{1}{y}}}}, that is,

\frac{1}{\log_{a}{e}}=\log_{e}{a}.

In the step before last, we have used the fact that the logarithm is a continuous function and that \lim_{y \rightarrow 0}{(1+y)^{1/y}}=e, while in the last step we have observed that (\log_{a}{e})^{-1}=\log_{e}{a} (Exercise).

More later,

Nalin Pithwa

 

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