4.10 Exponantiation and logarithm

For $0 < a \neq 1$ the function45

\begin{equation*} f(x) = a^x, \quad x\in D_f = \R, \end{equation*}

is called exponentiation of base $a$. This function extends the operation of elevation to power to non-integer exponents. For any real numbers $x$ and $y$, it applies the well known equality

\begin{equation*} a^x \cdot a^y = a^{x+y} \quad \text{a} \quad \big(a^x\big)^y = a^{xy}. \end{equation*}

In figure 4.14 the graph of the function $f$ is shown for different bases $a$.

Figure 4.14: Exponential functions.

In general, for $a > 1$ $f$ is strictly increasing (i.e. $f(x) < f(y)$ for any $x < y$), $D_f = \R$ and $H_f = (0,+\infty)$. For $a < 1$, $f$ is strictly decreasing (i.e. $f(x) > f(y)$ for any $x < y$), $D_f = \R$ and $H_f = (0,+\infty)$.

4.10.1 Logarithm

The logarithm is the inverse function of the exponentiation (only in the case of base different from one, otherwise the exponential function is not injective). More specifically, from the graph of the exponential function $f(x) = a^x$, $a \neq 1$, we see that for every real number $y$ there exists a real $x$ such that $a^x = y$. We say that a function with such a property is injective (in this case on the whole $\R$) and therefore invertible on its image set.The inverse function of the exponentiation of base $a$, $0 < a \neq 1$, is said logarithm of base $a$ and we denote it as $\log_a$. The domain of the exponentiation is the whole $\R$ and its image set is the interval $(0,+\infty)$. From this it follows that the domain of the logarithm, since it the function inverse of the exponentiation, is $D_{\log_a} = (0,+\infty)$ and its image set is $H_{\log_a} = \R$.

The reader has certainly already indirectly encountered logarithms through applications. For example the  Richter scale (that measures the intensity of earthquakes) or the  decibel scale (measuring the intensity of sound) are logarithmic.

Figure 4.15: Graph of some logarithmic functions with different bases.

Important properties of the logarithm can be derived from properties of the exponentiation:

\begin{equation}\label{eq-log-1}\tag{4.16} a^{\log_a x} = x, \quad x>0, \end{equation}

\begin{equation}\label{eq-log-2}\tag{4.17} \log_a a^x = x, \quad x\in\R, \end{equation}

\begin{equation}\label{eq-log-3}\tag{4.18} \log_a xy = \log_a x + \log_a y, \quad x,y > 0, \end{equation}

\begin{equation}\label{eq-log-4}\tag{4.19} \log_a x^y = y \log_a x, \quad x>0 \ \text{a} \ y\in\R. \end{equation}

Indeed, the first two equalities, (4.16) and (4.17), are merely an expression of the inverse relationship between the exponential and the logarithm, thus they apply by definition. Let us prove the equality (4.18). For positive $x,y$ there exist real $u,v$ such that

\begin{equation*} x = a^u \quad \text{a} \quad y = a^v. \end{equation*}

From this we have

\begin{equation*} xy = a^u \cdot a^v = a^{u+v}. \end{equation*}


\begin{equation*} \log_a xy = u + v = \log_a x + \log_a y. \end{equation*}

In a similar way, the property (4.19) can be proven.

Remark 4.3

The reader is certainly familiar with the operation called remove the logarithm. That is, saying the following: if

\begin{equation*} \log_a x = \log_a y, \end{equation*}

for some $x,y > 0$ and $0 < a \neq 1$, then

\begin{equation*} x = y. \end{equation*}

This operation is no magic. It is just about using the injectivity of the function $\log_a$. The same can be done with any injective function!

Question 4.11

Which is the domain of the function $f(x) = \log_a x^2$?

Show answer