#### Chapter 6 Elementary Functions

Section 6.4 Exponential and Logarithmic Functions

# 6.4.3 The Natural Exponential Function

There is a very special exponential function, sometimes also called the exponential function, that we will study now. In fact, all other exponential functions can be reduced to this special exponential function. It has Euler's number $e$ as its base. Its value is (approximately) equal to

$e=2.718281828459045235\dots .$

So, let us consider the graph of the exponential function - for the time being without any additional parameters -

$g:\mathrm{ }\left\{\begin{array}{ccc}\hfill ℝ& \hfill \to \hfill & \left(0;\infty \right)\hfill \\ \hfill x& \hfill ⟼\hfill & g\left(x\right)=e{}^{x}\hfill \end{array}$

which is, because of its base $e$, also called the $e$ function or natural exponential function:
Unsurprisingly, the natural exponential function shows the typical behaviour of exponential functions $x↦{a}^{x}$ $\left(a>1\right)$ already discussed in Section 6.4.2, after all we have only chosen a special value for the base, namely $a=e$. In particular, we note again that the natural exponential function is strictly increasing, i.e. for large negative values of $x$, it approaches the negative $x$-axis, and for $x=0$, it takes the value $1$.
##### Exercise 6.4.4
What does the graph of the function $h:ℝ\to \left(0;\infty \right)$, $x↦h\left(x\right)=e{}^{-x}$ look like, and which general properties has this function?

At the beginning of this subsection we claimed that the exponential functions described above can be reduced to the natural exponential function. This is done by means of the identity

${a}^{x}=e{}^{x·\mathrm{ln}\left(a\right)}$

that is valid for any real number $a>0$ and any real number $x$. Here, $\mathrm{ln}$ denotes the natural logarithmic function that will be studied in detail in the following Section 6.4.4.
##### Exercise 6.4.5
Explain why the identity ${a}^{x}=e{}^{x·\mathrm{ln}\left(a\right)}$ is valid.

In general natural exponential functions, the parameters ${f}_{0}$ and $\lambda$ occur that were already introduced in Section 6.4.2; thus, its functional description is as follows:

$f:\mathrm{ }\left\{\begin{array}{ccc}\hfill ℝ& \hfill \to \hfill & \left(0;\infty \right)\hfill \\ \hfill x& \hfill ⟼\hfill & f\left(x\right)={f}_{0}·e{}^{\lambda x}\hfill \end{array} .$

Again, the parameter ${f}_{0}$ describes initial values different from $1$, and the factor $\lambda$ in the exponent allows for different (positive or negative) growth rates. This shall be finally illustrated by means of an example.
##### Example 6.4.6
A series of experiments with radioactive iodine atoms (${}^{131}I$) results in the following mean data:
 Number of Iodine Atoms 10 000 5 000 2 500 1 250 etc. Number of Days Elapsed 0 8.04 16.08 24.12 etc.
In other words: every $8.04$ days the number of iodine atoms halves due to radioactive decay. For this reason one says in this context that the half-life $h$ of ${}^{131}I$ equals $8.04$ days.
The radioactive decay follows an exponential law:

$N\left(t\right)={N}_{0}·e{}^{\lambda t} .$

Our exponential function is here denoted by $N$; it describes the number of remaining iodine atoms. Accordingly, ${N}_{0}$ denotes the number of iodine atoms at the beginning, i.e. ${N}_{0}=10000$. The independent variable is in this case the time $t$ (measured in days). We expect the parameter $\lambda$ to be negative since the exponential function describes a decay process, i.e. a process with a negative growth rate. We will determine $\lambda$ from the measurement data.
After $h=8.04$ days only $5000$ iodine atoms are still present, i.e. $N\left(t=8.04\right)=5000=\frac{{N}_{0}}{2}$. Using the exponential law for the radioactive decay, we obtain:

$\frac{{N}_{0}}{2}={N}_{0}·e{}^{\lambda ·h} .$

Now, we can cancel ${N}_{0}$ on both sides of the equation and subsequently take the natural logarithm of the equation (see Section 6.4.4):

$\mathrm{ln}\left(\frac{1}{2}\right)=\mathrm{ln}\left(e{}^{\lambda ·h}\right) .$

We transform the left hand side according to the calculation rules for logarithmic functions (see Section 6.4.4): $\mathrm{ln}\left(1/2\right)=\mathrm{ln}\left(1\right)-\mathrm{ln}\left(2\right)=0-\mathrm{ln}\left(2\right)=-\mathrm{ln}\left(2\right)$. For the right hand side we note that the natural logarithmic function is the inverse function of the natural exponential function, i.e. $\mathrm{ln}\left(e{}^{\lambda ·h}\right)=\lambda ·h$; thus we have:

$\begin{array}{cccc}\hfill \hfill & \hfill -\mathrm{ln}\left(2\right)& \hfill =\hfill & \lambda ·h\hfill \\ \hfill ⇔\hfill & \hfill \lambda & \hfill =\hfill & -\frac{\mathrm{ln}\left(2\right)}{h} .\hfill \end{array}$

Inserting the half-life $h=8.04$ days of ${}^{131}I$ results in this case in

$\lambda \approx -0.0862\mathrm{ }\frac{1}{\text{day}} .$

Other radioactive substances have different half-lifes, e.g. ${}^{239}\mathrm{Pu}$ has a half-life of $24 000$ years, and hence they result in different values of the parameter $\lambda$ in the exponential law for the radioactive decay.