Logarithm

Introduction: logarithm as inverse of exponentiation

There are three simple ways equate a with b: using addition, a + c_diff = b, scaling, a * c_scaler = b or exponentiation, a^c = b. Similarly to the first case, where c_diff is called difference and is computed by subtraction, c_diff = b-a, and the second of multiplication, where factor c_scaler = b/a is computed by division (you say that subtraction and division are inverses of the addition and multiplication, correspondingly), the appropriate power c, called exponent, of base a is computed by the logarithm function, c = log_a(b). Therefore, the logarithm is a function, inverse of exponentiation.

Actually, because two arguments, the base a and power n, are not commutative in exponentiation aⁿ = b, there are two inverse operations. The first one, which computes a, given b and n, is known as n-th root. The logarithm, which computes n, given a and b is the other.

So, basically,

Formal definition

The logarithm is a function of two arguments, a and b, that computes such c

$c = \log_a b$

(2 )

that satisfies the equation

$a^c = b$

(1 )

The result of the log function is also called logarithm of b with respect to base a or exponent because it is the power that must be applied to the base a to get b.

Properties

The first thing we get by plugging (1 ) into (2 ):

$c = \log_a a^c.$

(3.1 )

It basically says that $\log_a a^c$ is equivalent to $c$ . Similar rule is obtained plugging (2 ) into (1 ):

$a^{\log_a b} = b$

(3.2 )

Because you must raise a into power 1 to get a, you may write

a¹ = a or

${\log_a a} = 1$

(4.1 )

Similarly, because to get 1 you must raise any number into power 0,

a^0 = 1

is equivalent to the logarithm property

${\log_a 1} = 0$

(4.2 )

and, finally,

a^{-\infty} = 0

is identical to

${\log_a 0} = -\infty$

(4.3 )

Note the controversy between (4.1 ) and (4.2 ): log₁1 = 1 according to the first rule and 0, according to the second.

Log (m*n) = log(m) + log(n)

Without loss of generality, we can define l and m to stand for powers of a,

\begin{array}{lcr} l = a^\lambda & \quad\overset\text{by 3.1}{\Longleftrightarrow}& \lambda = \log_a l\\ m = a^\mu & & \mu = \log_a m\end{array}

whereupon, the logarithm of the product is

\log_a (l \cdot m) = \log_a (a^\lambda \cdot a^\mu) = \log_a a^{\lambda + \mu} \overset{by\ 3.1}{=} \lambda + \mu \overset\text{by definition above}{=} \log_a l + \log_a m.

In short, we arrived at law log of a product equals the sum of the logarithmed operands:

$\log (l \cdot m) = \log l + \log m.$

(5 )

The base `a` is omitted because the law holds for any base.

This explains how logarithms replace multiplication with summation. They, further, replace exponentiation with multiplication:

Log (a^k) = k log(a)

Decomposing the first term m = m₁ * m₂ of the equation (5 ) above and using it two times, we can get

\log [(m_1 m_2) n] = \log (m_1 m_2) + \log n = \log m_1 + \log m_2 + \log n

You can see that this operation is associative

\log [m_1 (m_2 n)] = \log m_1 + (\log m_2 + \log n)

and commutative

\log (m n) = \log m + \log n = \log n + \log m = \log (n m)

The associativity and commutativity properties are valid for arbitrary number of terms

\log (m_1 m_2 \ldots m_n) = \log m_1 + \log m_2 + \cdots + \log m_n

One important case of having k identical items m₁ = m₂ = ... = m_n = m, gives:

$\log m^\textbf{k} = \textbf{k} \ \log m$

(11 )

If we are not familiar with logarithm of product, the same formula can be proven if we start with a¹=m, whence log_a(m)= 1 and, raised to the power, a^k=m^k gives log_a(m^k) = k = k * 1 = k log_a(m). The same can be demonstrated starting with identity (3.2 ), raised to the power k, $b^k = a^{k\log b}$ and, applying (1 ) with mapping $b^k \rightarrow b, k\log b \rightarrow c$ , we get the $\log b^k = k\log b$ .

Similarly, we can prove that factoring the power out of base, has the opposite effect:

$\log_{a^\textbf{k}} m = \textbf{1/k} \ \log_a m$

(12 )

Change of bases

Sometimes you have a library that computes logs in some fixed base, say log2, but you need to compute a logarithm of a number, say 3, in some different base, say 10. How do you compute x=log₁₀(3), given log2 function? You need to compute such x that 10^x = 3. Apply log2 to both left and right to get

\log_2 10^x = \log_2 3

Apply the last property of (11 )

x \log_2 10 = \log_2 3

and divide both sides by $\log_2 10$ and get

x = {\log_2 3 \over \log_2 10}

Generally,

\log_m n = {\log_a n \over \log_a m}

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