We shall now introduce a number, usually denoted by \(e\), which is of immense importance in higher mathematics. It is, like \(\pi\), one of the fundamental constants of analysis.

We define \(e\) as the number whose logarithm is \(1\). In other words \(e\) is defined by the equation \[1 = \int_{1}^{e} \frac{dt}{t}.\] Since \(\log x\) is an increasing function of \(x\), in the stricter sense of § 95, it can only pass once through the value \(1\). Hence our definition does in fact define one definite number.

Now \(\log xy = \log x + \log y\) and so \[\log x^{2} = 2\log x,\quad \log x^{3} = 3\log x,\ \dots,\quad \log x^{n} = n\log x,\] where \(n\) is any positive integer. Hence \[\log e^{n} = n\log e = n.\] Again, if \(p\) and \(q\) are any positive integers, and \(e^{p/q}\) denotes the positive \(q\)th root of \(e^{p}\), we have \[p = \log e^{p} = \log(e^{p/q})^{q} = q\log e^{p/q},\] so that \(\log e^{p/q} = p/q\). Thus, if \(y\) has any positive rational value, and \(e^{y}\) denotes the positive \(y\)th power of \(e\), we have \[\begin{equation*} \log e^{y} = y, \tag{1} \end{equation*}\] and \(\log e^{-y} = -\log e^{y} = -y\). Hence the equation (1) is true for all rational values of \(y\), positive or negative. In other words the equations \[\begin{equation*} y = \log x,\quad x = e^{y} \tag{2} \end{equation*}\] are consequences of one another so long as \(y\) is rational and \(e^{y}\) has its positive value. At present we have not given any definition of a power such as \(e^{y}\) in which the index is irrational, and the function \(e^{y}\) is defined for rational values of \(y\) only.

Example. Prove that \(2 < e < 3\). [In the first place it is evident that \[\int_{1}^{2} \frac{dt}{t} < 1,\] and so \(2 < e\). Also \[\int_{1}^{3} \frac{dt}{t} = \int_{1}^{2} \frac{dt}{t} + \int_{2}^{3} \frac{dt}{t} = \int_{0}^{1} \frac{du}{2 – u} + \int_{0}^{1} \frac{du}{2 + u} = 4\int_{0}^{1} \frac{du}{4 – u^{2}} > 1,\] so that \(e < 3\).]

$\leftarrow$ 202. Scales of infinity Main Page 204–206. The exponential function $\rightarrow$