Logarithm

1. Definition of a logarithm

Introductory example

How long has a capital stock of 1000.- to be invested with an interest rate of 5% such that the closing capital amounts 1980.-? The answer is:

$1000\cdot {{1.05}^{n}}\,\,=\,\,1980 \quad \Leftrightarrow \quad {{1.05}^{n}}\,\,=\,\,1.980$

Thus, we are looking for the exponent of 1.980 with basis 1.05, or the logarithm of 1.980 with basis 1.05.

n = logarithm of 1.980 with basis 1.05, abbreviated by $n\,\,=\,\,{{\log }_{1.05}}1.980$

Generalizin this example, we get the following definition:

Definition

$n\,\,=\,\,{{\log }_{a}}y \quad if \quad y\,=\,{{a}^{n}}$

Examples

${{\log }_{2}}8\,\,=\,\,3,\quad since \quad {{2}^{3}}=8 \quad \quad \quad {{\log }_{3}}81\,\,=\,\,4,\quad since \quad {{3}^{4}}=81$

${{\log }_{10}}1000\,\,=\,\,3,\,\,\, \quad since \quad {{10}^{3}}=1000 \quad \quad \quad {{\log }_{1.03}}1.0609\,\,=\,\,2, \quad since \quad {{1.03}^{2}}=1.0609$

${{\log }_{a}}1\,\,=\,\,0 \quad since \quad {{a}^{0}}\,\,=\,\,1$

2. Rules for logarithm

Introductory examples

Instead of $8\cdot 16\,\,=\,\,128 $ , we can write ${{2}^{3}}\cdot {{2}^{4}}\,\,=\,\,{{2}^{3+4}}\,\,\,=\,\,{{2}^{7}}$.

The exponent, i.e., the logarithm of the product is the sum of the exponents, i.e., the logarithm is the sum of the logarithms of the factors.

Instead of $64\,\,\div \,\,4\,\,=\,\,16 $ , we can write ${{2}^{6}}\,\,\div \,\,{{2}^{2}}\,\,=\,\,{{2}^{6-2}}\,\,=\,\,{{2}^{4}}$.

The exponent, i.e., the logarithm of the quotient, is the difference of the exponents, i.e., the logarithm of a quotient is the difference between the logarithm of the numerator and the logarithm of the denominator.

Instead of ${{32}^{4}}\,\,=\,\,1'048'576 $ , we can write ${{\left( {{2}^{5}} \right)}^{4}}\,\,=\,\,{{2}^{5\cdot 4}}\,\,=\,\,{{2}^{20}}$.

These examples demonstrate the following three rules for logarithms.

Three rules for logarithms:

$1. \quad {{\log }_{a}}u\cdot v\,\,=\,\,{{\log }_{a}}u\,\,+\,\,{{\log }_{a}}v$

$2. \quad {{\log }_{a}}\frac{u}{v}\,\,=\,\,{{\log }_{a}}u\,\,-\,\,{{\log }_{a}}v$

$3. \quad {{\log }_{a}}{{u}^{k}}\,\,=\,\,k\cdot {{\log }_{a}}u$

Examples

$1. \quad {{\log }_{2}}(32\cdot 64)\,\,=\,\,{{\log }_{2}}32\,\,+\,\,{{\log }_{2}}64\,\,=\,\,5+6\,\,=\,\,11$

$2. \quad {{\log }_{3}}({{9}^{4}})\,\,=\,\,4\cdot {{\log }_{3}}9\,\,=\,\,4\cdot 2\,\,=\,\,8$

$3. \quad {{\log }_{10}}(\frac{100}{1'000'000})\,\,=\,\,{{\log }_{10}}100\,\,-\,\,{{\log }_{10}}1'000'000\,\,=\,\,2-6=-4$

$4. \quad {{\log }_{2}}(\frac{1}{16})\,\,=\,\,{{\log }_{2}}1\,\,-\,\,{{\log }_{2}}16\,\,=\,\,0-4\,\,=\,\,-4$

3. Common logarithm and natural logarithm

In this section, we study logarithms with a special basis. For historical reasons, the basis 10 is important. The basis e (Euler number) is pivotal in financial market theory and the basis 2 is needed in computer sciences.

Electronic calculators usually compute at least one of these three logarithms.

A logarithm with an arbitrary basis can always be transformed into one of these special logarithms, and therefore any one of them allows us to determine logarithms of any basis.

Definition

${{\log }_{10}}y\,\,=\,\,\log y\,\,=\,\,\lg y\,\,\,\text{(common logarithm)}$

${{\log }_{e}}y\,\,=\,\,\ln y\,\,\,\text{(Logarithmus}\,\,\text{naturalis}\,\,\text{or}\,\,\text{natural logarithm)}$

${{\log }_{2}}y\,\,=\,\,lb\,y\,\,\,\,\,\text{(binary}\,\,\text{logarithm)}$

Examples

$\log 100\,\,=\,\,2 \quad \quad \log 0.001\,\,=\,\,-3$

$\log 2\,\,=\,\,0.301.. \quad \quad \log 4\,\,=\,\,0.602..$

The Euler number and the natural logarithm

Consider the following problem: 1 Fr. is invested for one year at an interest rate of 100% = 1. We now determine the future value of this investment when dividing that one year into different periods of equal length. The interest rate has, of course, to be adapted accordingly:

Period	Future value
1 year	${{K}_{1}}\,\,=\,\,(1+1)\,\,=\,\,2$
$\frac{1}{2}$ year	${{K}_{1}}\,\,=\,\,{{\left( 1+\frac{1}{2} \right)}^{2}}\,\,=\,\,{{1.5}^{2}}\,\,=\,\,2.25$
$\frac{1}{4}$ year	${{K}_{1}}\,\,=\,\,{{\left( 1+\frac{1}{4} \right)}^{4}}\,\,=\,\,{{1.25}^{4}}\,\,=\,\,2.44$
$\frac{1}{10}$ year	${{K}_{1}}\,\,=\,\,{{\left( 1+\frac{1}{10} \right)}^{10}}\,\,=\,\,{{1.1}^{10}}\,\,=\,\,2.59$
$\frac{1}{100}$ year	${{K}_{1}}\,\,=\,\,{{\left( 1+\frac{1}{100} \right)}^{100}}\,\,=\,\,{{1.01}^{100}}\,\,=\,\,2.70$
$\frac{1}{1\,mio.}$ year	${{K}_{1}}\,\,=\,\,{{\left( 1+\frac{1}{1\,mio.} \right)}^{1\,mio.}}\,\,=\,\,{{1.000'001}^{1'000'000}}\,\,=\,\,2.7182...$

Increasing the number of periods implies a proportionally shorter period for the adapted interest rate. The final outcome of this experiment is called continuous compounding: the compounded value increases, but there is a limit value.

In our example, the value of the investment of 1 Fr. compounds to 2.718281828459…. Fr. within a year. Thus, the capital increases by at most 171.828182..% with a nominal interest rate of 100%.

Note that e = 2.718281828459… is called the Euler number.

Moreover, ${{\log }_{e}}y\,\,=\,\,\ln y$ is called the natural logarithm, as mentioned above.

Examples

$\ln 100\,\,=\,\,4.605...\,\,\,since\,\,\,{{e}^{4.605..}}\,\,=\,\,100$

$\ln 1.1\,\,=\,\,0.0953...\,\,\,since\,\,\,{{e}^{0.0953...}}\,\,=\,\,1.1$

How can we calculate ${{\log }_{a}}y$ for an arbitrary basis a? It is easy to show that the following theorem holds:

Theorem for a change of a basis:

${{\log }_{a}}\,\,y=\,\,\frac{\log y}{\log a}\,\,=\,\,\frac{\ln y}{\ln a}$

Examples

$1. \quad {{\log }_{2}}8=\frac{\log 8}{\log 2}=\frac{\ln 8}{\ln 2}=3$

$2. \quad{{\log }_{3}}25=\frac{\ln 25}{\ln 3}=2.929947..$

Exercises

1)

$ {{\log }_{2}}256$

2)

$ {{\log }_{3}}2187$

3)

$ \log 200$

4)

$ \log 2000$

5)

$ \ln 200$

6)

Determine the maturity in the introductory example of section 1. "Definition of a logarithm".

Solution

An introduction to logarithm is available on the following website:

Math is fun

An introduction to logarithm is also available on Youtube:

Youtube

The rules for logarithm can be found on:

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