# 8.9. Generalised Exponential and Logarithm Functions

As ${e}^{\mathrm{X}}$ is the invers of ln we see that, according to the laws of logarithms,

${a}^{n}={e}^{\mathrm{ln}\left({a}^{n}\right)}={e}^{n\cdot \mathrm{ln}a}$

for all $a>0$ and $n\in ℤ$. This is an alternative way to calculate powers, a way however that we may well use as a guideline to introduce powers with arbitrary real exponents.

Definition:  Let $a,x\in ℝ$ be arbitrary with $a>0$. We set

 ${a}^{x}≔{e}^{x\cdot \mathrm{ln}a}$ [8.9.1]

As before we call a the base and x the exponent of the power  ${a}^{x}$. Being a value of the exponential function each power is strictly positive: ${a}^{x}>0$.

Note that, due to [8.8.25], we are allowed to use the notation ${a}^{x}=\mathrm{exp}\left(n\cdot \mathrm{ln}a\right)$ as well.

The new concept of powers is only applicable with a positive base. In this case, however we want to know if the new concept is a sequel to the common one. Thus we have to answer two questions:

• Do new and old values coincide for all $x\in ℚ$?

• Are the laws of exponents still valid?

We have a positive answer to both of them.

Proposition:  If $a>0$ we have for each $x\in ℚ$:

 ${a}^{x}=\mathrm{exp}\left(x\cdot \mathrm{ln}a\right)$ [8.9.2]

Proof:  Say $x=\frac{n}{m}$, $m>0$. Due to [8.8.2] and [8.7.6;10] we then have for ${a}^{x}$ in its old meaning:

${a}^{x}={a}^{\frac{n}{m}}=\sqrt[m]{{a}^{n}}=\mathrm{exp}\left(\mathrm{ln}\sqrt[m]{{a}^{n}}\right)=\mathrm{exp}\left(\frac{n}{m}\cdot \mathrm{ln}a\right)=\mathrm{exp}\left(x\cdot \mathrm{ln}a\right)$

We now turn to the laws of exponents. Besides the properties [8.8.2;3] we need the laws of logarithms [8.7.8;9] and the calculation rules for ${e}^{\mathrm{X}}$ [8.8.15;16].

Proposition (laws of exponents):  If $a,b>0$ the following properties hold for every $x,y\in ℝ$:

 1.   ${a}^{x}\cdot {b}^{x}={\left(a\cdot b\right)}^{x}$ $\frac{{a}^{x}}{{b}^{x}}=\left(\frac{a}{b}{\right)}^{x}$ $\frac{1}{{b}^{x}}=\left(\frac{1}{b}{\right)}^{x}$ [8.9.3] 2.   ${a}^{x}\cdot {a}^{y}={a}^{x+y}$ $\frac{{a}^{x}}{{a}^{y}}={a}^{x-y}$ $\frac{1}{{a}^{y}}={a}^{-y}$ [8.9.4] 3.   $\mathrm{ln}\phantom{\rule{0.1em}{0ex}}\left({a}^{x}\right)=x\cdot \mathrm{ln}a$ ${\left(\mathrm{exp}a\right)}^{x}=\mathrm{exp}\left(x\cdot a\right)$ [8.9.5] 4.   $\left({a}^{x}{\right)}^{y}={a}^{x\cdot y}$ [8.9.6]

Proof:

 1. ►  ${a}^{x}\cdot {b}^{x}={e}^{x\cdot \mathrm{ln}a}\cdot {e}^{x\cdot \mathrm{ln}b}={e}^{x\cdot \mathrm{ln}a+x\cdot \mathrm{ln}b}={e}^{x\cdot \left(\mathrm{ln}a+\mathrm{ln}b\right)}={e}^{x\cdot \mathrm{ln}\left(a\cdot b\right)}={\left(a\cdot b\right)}^{x}$ $\frac{{a}^{x}}{{b}^{x}}=\frac{{e}^{x\cdot \mathrm{ln}a}}{{e}^{x\cdot \mathrm{ln}b}}={e}^{x\cdot \mathrm{ln}a-x\cdot \mathrm{ln}b}={e}^{x\cdot \left(\mathrm{ln}a-\mathrm{ln}b\right)}={e}^{x\cdot \mathrm{ln}\frac{a}{b}}={\left(\frac{a}{b}\right)}^{x}$ The third equation is a special case of the second one. 2. ►  ${a}^{x}\cdot {a}^{y}={e}^{x\cdot \mathrm{ln}a}\cdot {e}^{y\cdot \mathrm{ln}a}={e}^{x\cdot \mathrm{ln}a+y\cdot \mathrm{ln}a}={e}^{\left(x+y\right)\cdot \mathrm{ln}a}={a}^{x+y}$ $\frac{{a}^{x}}{{a}^{y}}=\frac{{e}^{x\cdot \mathrm{ln}a}}{{e}^{y\cdot \mathrm{ln}a}}={e}^{x\cdot \mathrm{ln}a-y\cdot \mathrm{ln}a}={e}^{\left(x-y\right)\cdot \mathrm{ln}a}={a}^{x-y}$ Again, the third equation follows from the one before. 3. ►  $\mathrm{ln}\left({a}^{x}\right)=\mathrm{ln}\left({e}^{x\cdot \mathrm{ln}a}\right)=x\cdot \mathrm{ln}a$ ${\left(\mathrm{exp}a\right)}^{x}={e}^{x\cdot \mathrm{ln}\left(\mathrm{exp}a\right)}={e}^{x\cdot a}$ 4. ►  $\left({a}^{x}{\right)}^{y}={e}^{y\cdot \mathrm{ln}\left({a}^{x}\right)}\underset{{\phantom{\rule{0ex}{1em}}\phantom{\rule{0.1em}{0ex}}3\text{.}\phantom{\rule{0.1em}{0ex}}}}{=}{e}^{y\cdot x\cdot \mathrm{ln}a}={a}^{x\cdot y}$

In a first application of the new power concept we study exopnential equations, i.e. equations of the type

${a}^{x}=b$.

If $a\ne 1$ we get their unique solution by taking the logarithm on both sides.

Proposition and Definition:  For $a,b>0$, $a\ne 1$ the following equivalence holds

 ${a}^{x}=b\text{ }⇔\text{ }x=\frac{\mathrm{ln}b}{\mathrm{ln}a}$ [8.9.7]

The number ${\mathrm{log}}_{a}b≔\frac{\mathrm{ln}b}{\mathrm{ln}a}$ is called the logarithm to the base a of b. Apparently ${\mathrm{log}}_{a}b$ is that certain number that yields b if we take a to its power:

${a}^{{\mathrm{log}}_{a}b}=b$.

Proof:  We use [8.9.5] to get:  ${a}^{x}=b\text{ }⇔\text{ }\mathrm{ln}\phantom{\rule{0.1em}{0ex}}{a}^{x}=\mathrm{ln}b\text{ }⇔\text{ }x\cdot \mathrm{ln}a=\mathrm{ln}b\text{ }⇔\text{ }x=\frac{\mathrm{ln}b}{\mathrm{ln}a}$.

The logarithms to a fixed base a allow to introduce generalised logarithm functions.

Definition:  For any $a>0$, $a\ne 1$ we call the function

 ${\mathrm{log}}_{a}:{ℝ}^{>0}\to ℝ$ [8.9.8]

the (generalised) logarithm function to the base a. Instead of ${\mathrm{log}}_{a}\left(x\right)$ we often write ${\mathrm{log}}_{a}x$ which is more common. Also, we note that every logarithm function is just a multiple of ln:  ${\mathrm{log}}_{a}=\frac{1}{\mathrm{ln}a}\cdot \mathrm{ln}$.

With $\mathrm{ln}e=1$ we have ${\mathrm{log}}_{e}=\mathrm{ln}$, thus the logarithm to the base e is the natural logarithm. Two further logarithm functions have names for their own:

• the common logarithm  $\mathrm{lg}≔{\mathrm{log}}_{10}$

• the binary (or dual) logarithm  $\mathrm{ld}≔{\mathrm{log}}_{2}$

As the logarithms are multiples of ln they take over most of its properties. The laws of logarithms [8.7.6-10] for example analogously hold for ${\mathrm{log}}_{a}$ as well. Furthermore, ${\mathrm{log}}_{a}$ is integrable and arbitrary often differentiable. Its graph is the result of a perpendicular dilation of that of ln:

Now that the concept of power has been extended new functions could be introduced. Generalised power functions are our first example.

Definition:  For any real a the function

 ${\mathrm{X}}^{a}≔{e}^{a\cdot \mathrm{ln}}:{ℝ}^{>0}\to ℝ$ [8.9.9]

is called the (generalised) power function with exponent a. Obviously ${\mathrm{X}}^{a}\left(x\right)={e}^{a\cdot \mathrm{ln}x}={x}^{a}$.

Power functions are differentiable and integrable. Derivatives and primitives follow the "common scheme":

Proposition:  Each power function ${\mathrm{X}}^{a}$ is

 1.   differentiable with $\left({\mathrm{X}}^{a}{\right)}^{\prime }=a\phantom{\rule{0.1em}{0ex}}{\mathrm{X}}^{a-1}$ [8.9.10] 2.   arbitrary differentiable and $\left({\mathrm{X}}^{a}{\right)}^{\left(n\right)}=\prod _{i=0}^{n-1}\left(a-i\right)\cdot {\mathrm{X}}^{a-n}$  for all $n>0$ [8.9.11] 3.   integrable and $\frac{1}{a+1}{\mathrm{X}}^{a+1}$ is a primitive of ${\mathrm{X}}^{a}$ if $a\ne -1$. [8.9.12]

Proof:

1.   Differentiability is due to the chain rule [7.7.8] which also provides the derivative:

$\left({\mathrm{X}}^{a}{\right)}^{\prime }=\left({e}^{\mathrm{X}}\circ a\cdot \mathrm{ln}{\right)}^{\prime }=\left(\left({e}^{\mathrm{X}}{\right)}^{\prime }\circ a\cdot \mathrm{ln}\right)\cdot a\cdot {\mathrm{ln}}^{\prime }=\left({e}^{\mathrm{X}}\circ a\cdot \mathrm{ln}\right)\cdot a\cdot {\mathrm{X}}^{-1}={\mathrm{X}}^{a}\cdot a\cdot {\mathrm{X}}^{-1}=a\phantom{\rule{0.1em}{0ex}}{\mathrm{X}}^{a-1}$

2.   The proof is by induction with the base step already done in 1. Thus assume ${\mathrm{X}}^{a}$ is n-times differentiable and the derivative formula [8.9.11] is valid. Note that the nth derivative $\left({\mathrm{X}}^{a}{\right)}^{\left(n\right)}=\prod _{i=0}^{n-1}\left(a-i\right)\cdot {\mathrm{X}}^{a-n}$ is a multiple of a power function and thus differentiable as well with

${\left({\mathrm{X}}^{a}\right)}^{\left(n+1\right)}=\prod _{i=0}^{n-1}\left(a-i\right)\cdot \left({\mathrm{X}}^{a-n}{\right)}^{\prime }=\prod _{i=0}^{n-1}\left(a-i\right)\cdot \left(a-n\right)\cdot {\mathrm{X}}^{a-n-1}=\prod _{i=0}^{n}\left(a-i\right)\cdot {\mathrm{X}}^{a-\left(n+1\right)}$

3.   The case $a=-1$ is well-known. If $a\ne -1$ we may differentiate the function $\frac{1}{a+1}{\mathrm{X}}^{a+1}$ according to 1. and get ${\mathrm{X}}^{a}$ as its derivative.

The generalised exponential functions are another application of the extended power concept.

Definition:  For each $a>0$ the function

 ${a}^{\mathrm{X}}≔{e}^{\mathrm{X}\cdot \mathrm{ln}a}:ℝ\to {ℝ}^{>0}$ [8.9.13]

is called the (generalised) exponential function with base a. Its values are ${a}^{\mathrm{X}}\left(x\right)={e}^{x\cdot \mathrm{ln}a}={a}^{x}$.

Two of the exponential functions are not new for us:

• The exponential function with base 1 is the constant function 1, actually because $\mathrm{ln}1=0$:

${1}^{\mathrm{X}}={e}^{\mathrm{X}\cdot \mathrm{ln}1}={e}^{0}=1$

• The exponential function with base e is the natural exponential function ${e}^{\mathrm{X}}=\mathrm{exp}$. This is due to $\mathrm{ln}e=1$:

${e}^{\mathrm{X}}=\mathrm{exp}\circ \left(\mathrm{X}\cdot \mathrm{ln}e\right)=\mathrm{exp}\circ \mathrm{X}=\mathrm{exp}$

This identity now justifies the power notation for the exponential function in terms of content: The natural exponential function is a special exponential function, namely that with base e. Thus the notation introduced in 8.8 is more than a symbolic one.

As the values of exponential functions are powers the laws of exponents [8.9.4-6] apply and thus provide respective rules for these functions. As an example we we note

${a}^{\mathrm{X}}\left(x+1\right)={a}^{\mathrm{X}}\left(x\right)\cdot {a}^{\mathrm{X}}\left(1\right)=a\cdot {a}^{\mathrm{X}}\left(x\right)$

as a special case of [8.9.4], which means: If we increment x by one unit the value turns to the a-fold one.

The inner function of the decomposition ${a}^{\mathrm{X}}={e}^{\mathrm{X}}\circ \mathrm{X}\cdot \mathrm{ln}a$ is a multiple of X. The graph of an exponential function thus is the result of a horizontal dilation of that of ${e}^{\mathrm{X}}$:

Exponential functions are differentiable and integrable. Calculating their derivatives and primitives is an easy task.

Proposition:  Every exponential function ${a}^{\mathrm{X}}$ is

 1.   differentiable with $\left({a}^{\mathrm{X}}{\right)}^{\prime }=\mathrm{ln}a\cdot {a}^{\mathrm{X}}$ [8.9.14] 2.   arbitrary often differentiable and $\left({a}^{\mathrm{X}}{\right)}^{\left(n\right)}={\left(\mathrm{ln}a\right)}^{n}\cdot {a}^{\mathrm{X}}$ for all $n>0$ [8.9.15] 2.   integrable and $\frac{1}{\mathrm{ln}a}{a}^{\mathrm{X}}$ is a primitive of ${a}^{\mathrm{X}}$ if $a\ne 1$ [8.9.16]

Proof:

1.   Again, differentiability and derivative formula are due to the chain rule [7.7.8]:

$\left({a}^{\mathrm{X}}{\right)}^{\prime }=\left({e}^{\mathrm{X}}\circ \mathrm{X}\cdot \mathrm{ln}a{\right)}^{\prime }=\left(\left({e}^{\mathrm{X}}{\right)}^{\prime }\circ \mathrm{X}\cdot \mathrm{ln}a\right)\cdot \left(\mathrm{X}\cdot \mathrm{ln}a{\right)}^{\prime }=\left({e}^{\mathrm{X}}\circ \mathrm{X}\cdot \mathrm{ln}a\right)\cdot \mathrm{ln}a=\mathrm{ln}a\cdot {a}^{\mathrm{X}}$

2.   This is a straightforward proof by induction with the base step already shown by 1. For the induction step we just note that the nth derivative of ${a}^{\mathrm{X}}$ is a multiple of ${a}^{\mathrm{X}}$ and therefor differentiable. Its derivative $\left({a}^{\mathrm{X}}{\right)}^{\left(n\right)}$ just adds the factor $\mathrm{ln}a$ another time so that the derivative formula is valid for $\left({a}^{\mathrm{X}}{\right)}^{\left(n+1\right)}$ as well.

3.   The case ${1}^{\mathrm{X}}=1$ is trivial and if $a\ne 1$ we just need to differentiate the function $\frac{1}{\mathrm{ln}a}{a}^{\mathrm{X}}$ which is easily done using 1.

${e}^{\mathrm{X}}$ and ln are inverse functions to each other. This is also true for generalised exponential and logarithm functions.

Proposition:  If $a>0$, $a\ne 1$ then ${a}^{\mathrm{X}}$ and ${\mathrm{log}}_{a}$ are invers to each other:

 $\begin{array}{l}{a}^{\mathrm{X}}\circ {\mathrm{log}}_{a}=\mathrm{X}|{ℝ}^{>0}\hfill \\ {\mathrm{log}}_{a}\circ {a}^{\mathrm{X}}=\mathrm{X}\hfill \end{array}$ [8.9.17]

Proof:  We only need to show that both functions cancel out each other:

1. The identity ${a}^{{\mathrm{log}}_{a}x}=x$ is valid for all $x>0$ according to [8.9.7].

2. ${\mathrm{log}}_{a}\left({a}^{x}\right)=\frac{1}{\mathrm{ln}a}\cdot \mathrm{ln}\left({a}^{x}\right)=\frac{1}{\mathrm{ln}a}\cdot x\cdot \mathrm{ln}a=x$ holds for all x due to [8.9.8] and [8.9.5]. 