The logarithm arises as the inverse operation of exponentiation. Whereas a power allows one to determine a result once the base and the exponent are known, the logarithm allows one to determine the exponent once the base and the result are known.
More precisely, given a base \(b>0\), with \(b\neq 1\), and a number \(x>0\), the logarithm of \(x\) to base \(b\), denoted by \(\log_b(x)\), is the exponent to which \(b\) must be raised in order to obtain \(x\). In symbols,
\[ \log_b(x)=y \iff b^y=x. \]
Starting from this definition, we shall prove the main properties of logarithms, which allow products, quotients and powers to be transformed into simpler expressions. We shall also establish the fundamental identities linking logarithms and exponentials, together with the change-of-base formula.
Contents
- Definition of the logarithm
- Fundamental identities and special values
- Logarithm of a product
- Logarithm of a quotient
- Logarithm of a power
- Change-of-base formula
Definition of the logarithm
Let \(b\) and \(x\) be real numbers such that
\[ b>0,\qquad b\neq 1,\qquad x>0. \]
The logarithm of \(x\) to base \(b\) is defined to be the unique real number \(y\) such that
\[ b^y=x. \]
In symbols,
\[ \log_b(x)=y \iff b^y=x. \]
The base of the logarithm must be positive and different from \(1\), while its argument must be strictly positive.
The condition \(b>0\) guarantees that the power \(b^y\) is defined for every real exponent \(y\). The condition \(b\neq 1\) is necessary because
\[ 1^y=1 \]
for every \(y\in\mathbb{R}\). Therefore, when \(b=1\), no exponent produces a positive number different from \(1\), while every real exponent produces \(1\). Consequently, the logarithm cannot be defined as an inverse operation.
Moreover, since
\[ b^y>0 \]
for every \(y\in\mathbb{R}\), the logarithm is defined only for positive arguments.
For instance,
\[ \log_2(8)=3, \]
since
\[ 2^3=8. \]
Likewise,
\[ \log_3\left(\frac{1}{9}\right)=-2, \]
since
\[ 3^{-2}=\frac{1}{9}. \]
The logarithm may therefore take positive, zero or negative values, whereas its argument must always be positive.
Fundamental identities and special values
Since the logarithm and exponentiation are inverse operations, the following fundamental identities hold:
\[ b^{\log_b(x)}=x, \qquad x>0, \]
and
\[ \log_b(b^y)=y, \qquad y\in\mathbb{R}, \]
where \(b>0\) and \(b\neq 1\).
The first identity follows immediately from the definition of the logarithm: \(\log_b(x)\) is, by definition, the exponent to which \(b\) must be raised in order to obtain \(x\).
The second expresses the same relationship in the reverse direction: the logarithm to base \(b\) of the power \(b^y\) returns the exponent \(y\).
The definition also yields two important special values:
\[ \log_b(1)=0 \]
since
\[ b^0=1, \]
and
\[ \log_b(b)=1 \]
since
\[ b^1=b. \]
For instance,
\[ 5^{\log_5(12)}=12 \]
and
\[ \log_2(2^{-3})=-3. \]
Logarithm of a product
The logarithm of a product is equal to the sum of the logarithms of the factors. If
\[ x>0,\qquad y>0, \]
then
\[ \log_b(xy)=\log_b(x)+\log_b(y), \]
where \(b>0\) and \(b\neq 1\).
Proof. Set
\[ u=\log_b(x) \qquad\text{and}\qquad v=\log_b(y). \]
By the definition of the logarithm,
\[ x=b^u \qquad\text{and}\qquad y=b^v. \]
Multiplying the two equalities, we obtain
\[ xy=b^u\cdot b^v=b^{u+v}. \]
Taking the logarithm to base \(b\) of both sides,
\[ \log_b(xy)=\log_b\left(b^{u+v}\right)=u+v. \]
Since \(u=\log_b(x)\) and \(v=\log_b(y)\), it follows that
\[ \log_b(xy)=\log_b(x)+\log_b(y). \]
Example. Let us compute
\[ \log_2(4\cdot 8). \]
Applying the product property,
\[ \log_2(4\cdot 8)=\log_2(4)+\log_2(8)=2+3=5. \]
Indeed,
\[ 4\cdot 8=32 \qquad\text{and}\qquad \log_2(32)=5. \]
The property extends to a product of any finite number of positive factors. If \(n\in\mathbb{N}\), \(n\geq 2\), and \(x_1,\ldots,x_n>0\), then
\[ \log_b(x_1x_2\cdots x_n) = \log_b(x_1)+\log_b(x_2)+\cdots+\log_b(x_n). \]
Logarithm of a quotient
The logarithm of a quotient is equal to the difference between the logarithm of the numerator and the logarithm of the denominator. If
\[ x>0,\qquad y>0, \]
then
\[ \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y), \]
where \(b>0\) and \(b\neq 1\).
Proof. Set
\[ u=\log_b(x) \qquad\text{and}\qquad v=\log_b(y). \]
By the definition of the logarithm,
\[ x=b^u \qquad\text{and}\qquad y=b^v. \]
Dividing the first equality by the second, we obtain
\[ \frac{x}{y}=\frac{b^u}{b^v}=b^{u-v}. \]
Taking the logarithm to base \(b\) of both sides,
\[ \log_b\left(\frac{x}{y}\right) = \log_b\left(b^{u-v}\right) = u-v. \]
Since \(u=\log_b(x)\) and \(v=\log_b(y)\), it follows that
\[ \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y). \]
Example. Let us compute
\[ \log_3\left(\frac{81}{9}\right). \]
Applying the quotient property,
\[ \log_3\left(\frac{81}{9}\right) = \log_3(81)-\log_3(9) = 4-2 = 2. \]
Indeed,
\[ \frac{81}{9}=9 \qquad\text{and}\qquad \log_3(9)=2. \]
Logarithm of a power
The logarithm of a power is equal to the exponent multiplied by the logarithm of the quantity being raised to that power. If
\[ x>0 \]
and \(\alpha\in\mathbb{R}\), then
\[ \log_b(x^\alpha)=\alpha\log_b(x), \]
where \(b>0\) and \(b\neq 1\).
Proof. Set
\[ u=\log_b(x). \]
By the definition of the logarithm,
\[ x=b^u. \]
Raising both sides to the power \(\alpha\), we obtain
\[ x^\alpha=(b^u)^\alpha=b^{\alpha u}. \]
Taking the logarithm to base \(b\) of both sides,
\[ \log_b(x^\alpha) = \log_b\left(b^{\alpha u}\right) = \alpha u. \]
Since \(u=\log_b(x)\), it follows that
\[ \log_b(x^\alpha)=\alpha\log_b(x). \]
Example. Let us compute
\[ \log_2(8^4). \]
Applying the power property,
\[ \log_2(8^4)=4\log_2(8)=4\cdot 3=12. \]
Indeed,
\[ 8^4=(2^3)^4=2^{12}, \]
and therefore
\[ \log_2(8^4)=12. \]
An important particular case is obtained by setting \(\alpha=-1\):
\[ \log_b\left(\frac{1}{x}\right) = \log_b(x^{-1}) = -\log_b(x), \qquad x>0. \]
Likewise, setting \(\alpha=\displaystyle\frac{1}{n}\), with \(n\in\mathbb{N}\) and \(n\geq 2\), we obtain
\[ \log_b\left(\sqrt[n]{x}\right) = \frac{1}{n}\log_b(x), \qquad x>0. \]
Change-of-base formula
The change-of-base formula allows a logarithm to a given base to be expressed in terms of logarithms computed in another base. If
\[ x>0,\qquad b>0,\qquad b\neq 1,\qquad c>0,\qquad c\neq 1, \]
then
\[ \log_b(x)=\frac{\log_c(x)}{\log_c(b)}. \]
Proof. Set
\[ y=\log_b(x). \]
By the definition of the logarithm,
\[ b^y=x. \]
Taking the logarithm to base \(c\) of both sides, we obtain
\[ \log_c(b^y)=\log_c(x). \]
By the power property,
\[ y\log_c(b)=\log_c(x). \]
Since \(b\neq 1\), we have \(\log_c(b)\neq 0\). We may therefore divide both sides by \(\log_c(b)\):
\[ y=\frac{\log_c(x)}{\log_c(b)}. \]
Recalling that \(y=\log_b(x)\), it follows that
\[ \log_b(x)=\frac{\log_c(x)}{\log_c(b)}. \]
Common and natural logarithms
Two particularly important bases are \(10\) and \(e\). The logarithm to base \(10\) is called the common logarithm and, in elementary mathematics, is often denoted by
\[ \log(x)=\log_{10}(x), \]
while the logarithm to base \(e\) is called the natural logarithm and is denoted by
\[ \ln(x)=\log_e(x). \]
The change-of-base formula therefore allows us to write
\[ \log_b(x)=\frac{\ln(x)}{\ln(b)} \]
or
\[ \log_b(x)=\frac{\log(x)}{\log(b)}. \]
Example. Let us express \(\log_2(40)\) in terms of the natural logarithm:
\[ \log_2(40)=\frac{\ln(40)}{\ln(2)}. \]
Using a calculator, we obtain
\[ \log_2(40)\approx 5.322. \]
This value lies between \(5\) and \(6\), as expected from
\[ 2^5=32<40<64=2^6. \]