## 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) 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|, $a^{x_{0}-\varepsilon} \leq y_{0}-\delta , or

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

$g(y_{0})-\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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