The Logarithm and the Exponential

If you have a number on an interval of the Real numbers, $x \in X \subset \mathbb{R}$, and have in an image space of the exponential function a mapping from such element to a nesting in the argument of the exponential, $f:x\mapsto b^x$, where $b$ is a constant, the base of the exponent, and $x$ is the variable argument to $f$ the exponential, and $y=f(x)\in Y\subset\mathbb{R}^+$, the image or ordinate. For example, for $b=2, X=(-1,1)$, then $Y=(0.5,2)$. The logarithm takes a number $y\in Y \subset \mathbb{R}^+$, where $Y$ is the domain for the logarithm, which is a subset of the Real numbers having a lower bound of zero and no upper bound. The logarithm returns the exponent $x$ such that $b^x=y$, or $x=\log_b(y)$, so that the logarithm of the exponential function with matching base is the identity function, $\log_b(b^x)=x$, and the exponential of the logarithm with matching base is the identity function, $b^{\log_b(y)}=y$. The nature of $\log_c(b^x), c\ne b$ is derived below, but basically it's only off from the identity by a constant factor.

$$ \log_2(16) = 4, \quad 2^4 = 16 $$ $$ \log_{2}(0.5) = -1, \quad 2^{-1} = 0.5 $$

There is one base which is special, called the Natural exponent base, $e$, which is to seven decimal places $2.718282$, and the logarithm with that base is called the Natural logarithm, $\ln(x)=\log_e(x)$.

All image points of logarithms with argument values greater than zero and less than one are negative real numbers, regardless of base, $\log(x)\lt 0:x\in (0,1)$, since the exponent required to produce a proper fraction is negative, $0\lt e^x\lt 1: x\lt 0$. All logarithms of values between one and the base itself return values which are proper fractions, since the exponent of a $b$-based exponential required to produce a number in the $(1,b)$ range is always a fraction between zero and one. This should be clear from the Zero Exponent article, since $b^0=1$, and $b^1=b$. Finally, or perhaps the beginning of the study of things like square-root of negative numbers, as the logarithm of a negative number is not defined in the real number system, but is fundamental to complex analysis as a complex number ($a+bi:a,b\in \mathbb{R}$), symbolized with $i$ (when it's not an index, as in multivariable calculus subscripts). Complex analysis studies properties of functions of complex variables, involving $i=\sqrt{-1}$ in domain and image, whose declaration as such entails its arithmetic, being anything $i^2=-1$, such as $-i$, which defines the imaginary dimension. But more on the appearance of the complex after Taylor series, for now the regular pre-calc treatment of logarithms and exponential functions.

The identities of the logarithm are implied by the behavior of the exponential functions, which we know arithmetically. They are for exponentials, $y=b^{x+w}=b^x b^w$, and $y=b^{x-w}=b^x/b^w$, which are direct result of the definition of real exponent as being factors of $b$, and possibly a factor of $b^r\ge 1, 0\le r \le 1$. So, for logarithms they are, $y=\log_b(xw)=\log_b(x) + \log_b(w)$, and $y=\log_b(x/w)=\log_b(x) - \log_b(w)$, which take some getting used to thinking in terms of inverse space, but are pretty simple to derive, if not remember, and very easy to use. The logarithm power rule is proven by inverting the equation into an exponential, and then using the exponential power rule, which is, $(b^x)^n=b^{xn}$.

\begin{equation}\label{log_power_rule} y=\log_b(c^x) \rightarrow c^x=b^y \rightarrow c=b^{y/x} \rightarrow \log_b(c)=\frac{y}{x} \rightarrow y=x\log_b(c) \end{equation}

The logarithm power rule, eq. \eqref{log_power_rule}, states that the logarithm of an exponential function, regardless of the base of the log or the base of the exponential in its argument, is equal to the value formerly-in-exponent times the logarithm of the base, $\log_b(c^x)=x\log_b(c), \forall\, b, c \gt 0$.

And just as there's no simplification for the addition of two different exponentials, there is no simplification for adding two different arguments of a logarithm.

The exponential function is positive for all numbers on the real number line, zero for negative infinity, unbounded from above, and smooth and continuous. Since the exponential is always greater than zero for any point in the domain of reals, the logarithm is imaginary (has a factor of square root of $-1$) for all domain points less than zero, so until you need it, like in quantum mechanics, it's only basically defined for positive numbers. The following fundamental examples show how these related functions work, starting with the formulas for changing from one base (including to/from the natural base, $e$), for both the exponential and the logarithm functions (of domain and image).

\begin{equation}\label{exp_change_of_base} y=b^x \rightarrow x\log_c(b)=\log_c(y) \rightarrow b^x = c^{x\log_c(b)} \end{equation}

In these steps, we declared the image point in terms of the exponential, base $b$, of the exponential's domain point, $x$. Then in order to use the log power rule, base $c$, we took that log of both sides, applied the rule, and used the fact that $c^{\log_c(y)}=y$.

\begin{equation}\label{log_change_of_base} y=\log_b(x) \rightarrow x=b^y \rightarrow \log_c(x) = y\log_c(b) \rightarrow \log_b(x) = \frac{\log_c(x)}{\log_c(b)} \end{equation}

Where the first arrow represents the use of the definition of the logarithm, the second is that of taking the natural logarithm of both sides and then applying the power rule, eq. \eqref{log_power_rule}, and the last is solving for $y$ and asserting the premise.

The Natural Exponent, Historically

The exponential function is a function of the form $f(x)=b^x$, where $b$ is a constant, the base of the exponent, and $x$ is the variable exponent. For real-$x$, this is just the quantity produced by sequentially multiplying the base a floored value of $x$ number of times, $x_{fl}:=\text{Floor}(x)= \lfloor x\rfloor$, which is defined to be the greatest integer less than or equal to $x$, with a factor of base to the power, $x_{rem}=x-\lfloor x\rfloor$, where the "remainder" factor, $1\le b^{x_{rem}} \lt b$, which you should recall from the Zero Exponent article.

$$ \lfloor -6.5 \rfloor = -7, \quad \lfloor 6.5 \rfloor = 6, \quad $$

In practice, the base is always a special irrational number, originally discovered by Jacob Bernoulli (1683), to be the limit of $(1+1/n)^n$ as $n$ approaches infinity, which is $2.718282\ldots$.

This is a plot of the product of a unit-sized initial investment with compounding interest rate of hyperbolically variable apr, which is the inverse-number-of-times-compounded in the unit-duration investment period, where number of times compounded is understood to be this variable, $x$, in $(1+1/x)^x$ on the interval $[1,100]$, Legend: blue $=x*1$, pistachio $=x*10$, 3rd color $=x*100$, + red $=e$.

So, for $x$-large, the expression $(1+1/x)^x$ approaches the number $e$, from below, so if we take the logarithm of both sides, we have a first order approximation to the logarithm of $e$, which approaches $1$, also from below.

$$ \ln(1+\epsilon) \approx \epsilon \quad \text{for } \epsilon\ll 1 $$

This is a plot of the natural logarithm of $(1+1/x)$ (blue) and the function $1/x$ (red) on the interval $[1,30]$, showing that for large $x$, $\ln(1+1/x)$ approaches $1/x$, since $x$ only appears as inverse-$x$, we're really looking at small parameter $\lim_{\epsilon\ll 1}\epsilon\approx \ln(1+\epsilon)$. There is a highlighted point at $x=1$, where $\ln(2)\approx 0.693147\ldots$.

The logarithm was revealed by Euler to be the inverse of the exponential function (1748). Given a variable $y$ related to $x$ by its logarithm (and $x$ being any number, positive, negative, whole number, or in-between), $x=b^y$ is the same statement as $y=\log_b(x)$ . For exponential-base $2$, it can be stated like so:

$$ y=\log_2(x) \rightarrow x=2^y $$

Likewise, the Natural logarithm is the inverse of the Natural-based exponent.

The Natural exponential function is also encountered in Statistical mechanics (pioneered by American physicist Josiah Gibbs, after the Civil war), where it is found with an exponent of energy, as opposed to $x$-alone, the identity function.

The irrational number $e$, $2.718282\ldots$, can be calculated in myriad ways ( [Dunham] ) one of which has been demonstrated with a graphical plot. It was posited by Leonhard Euler that the natural base can be found by solving the equation $b^x=1+x$, where one orients with the equation first by noting that both left and right are easily satisfied for $x=0$. For points very close to $x=0$, regardless of the base $b$, the curve is approximately a line, of varying slope, all passing through point $(0,1)$: $$ b^x=1+kx $$ Because we know $b^0=1$ (regardless of $b$, more at zero-exponent ), the curves of different bases go through the point $(0,1)$, and it turns out they have different slopes, with the "natural" base having a slope of unity ($k=1$), otherwise $k=\ln(b)$.

Below are plots of different bases (values of $b$ constant), with lines drawn at slopes $+0.1$ greater slope, and $-0.1$ less slope than the slope of the curve at the point $(0,1)$, to demonstrate a behavior common to all exponentials regardless of the charactaristics of the function, $f(x)$, in the exponent position ($e^{f(x)}$). By looking at the plot of $b^x$, for some constant $b \in \{2, e, 3\}$, and drawing a line in the right place (tangent to the exponential curve, at point $f(x)=x=0$), we exercise one way. First, inspect a plot of the exponential function with base $b=2$, below.

This plot of the exponential with base of $2$ (light pistachio color), on the interval $-2$ to $2$ has a red line drawn through point (0,1) at a slope of 0.6 and a similar blue line at slope 0.8, with the bisection indicating $k=0.7 = \ln(2)$.

All the curves are stroked the same width, so looking at the triple crossing, drawn at the exponent of zero, point (0,1), one can see that the red, and blue lines have slopes just a little less and a little more than the slope of the exponential curve at that point.

This plot of the "natural base" of $~2.718$ , from $-2$ to $2$ (pistachio) has a red line drawn through point (0,1) at a slope of 0.9 and a similar blue line at slope 1.1, indicating the "natural", ideal, slope of one.

This plot of the base of $3$ from $-2$ to $2$ (pistachio) has a red line drawn through point (0,1) at a slope of 1.0 and a similar blue line at slope 1.2, indicating $k=1.1=\ln(3)$.

This plot of the base of $3$ from $(-0.0002)$ to $0.0002$ (pistachio) has a red line drawn through point (0,1) at a slope of 1.0 and a similar blue line at slope 1.2, indicating a slope of 1.1 for the exponential curve (pistachio), when it is very close to the origin.

Thus the slope of the exponential curve at the origin is equal to the natural logarithm of the base, $\ln(3)$, which is approximately $1.0986$, graphically demonstrated.


1. Scriba, Christoph J. (1963). The inverse method of tangents: A dialogue between Leibniz and Newton. Archive for History of Exact Sciences 2 (2):113-137.


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