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Common and Natural Logarithms

what you'll learn...

overview

•  common logarithm or logarithm of base $10$$10$

•  natural logarithm or logarithm of base $e$$e$

The number $e$$e$ is also introduced in a thought-provoking and ingenious way.

base 10

In decimal system, the numbers are represented in the base $10$$10$

The place value is in units, tens, hundreds, etc. This is in powers of $10$$10$.

We learned that

•  logarithm can be specified in any base. eg: ${\mathrm{log}}_{2}3,{\mathrm{log}}_{10}3$${\log}_{2} 3 , {\log}_{10} 3$

•  a logarithm of one base can be converted to logarithm of another base. eg: ${\mathrm{log}}_{2}3={\mathrm{log}}_{7}3÷{\mathrm{log}}_{7}2$${\log}_{2} 3 = {\log}_{7} 3 \div {\log}_{7} 2$ (which will be explained later)

Knowing these, the base of logarithm can be standardized to a numerical value for uniformity and simplicity.

One of the good choices is "log of base $10$$10$, as base of decimals is $10$$10$".

common or natural

Logarithm of base $10$$10$ is called common logarithm.

If the logarithm is specified as $\mathrm{log}3$$\log 3$ (without the base), then it represents logarithm to the base $10$$10$.

In coming up with a standard for base of logarithms, the considerations were only related to being uniform with the base of decimal number system. The application scenarios and the practical uses of logarithm were not taken in consideration.

Let us examine an application scenario and explore an use of logarithm. This discussion will make "natural logarithm" evident.

The logic is little involved. Students are advised to go through this.

linear and proportional

Consider a medicine being absorbed by cells in body.

Consider $100mg$$100 m g$ of medicine in the blood. Let us examine two mechanisms (out of several under different levels and environment) by which the medicine is absorbed.

In the first case, a constant amount of medicine comes in contact with cells and is absorbed. $20mg$$20 m g$ of medicine comes in contact with the cells in $1$$1$ hour and so every $1$$1$ hour $20mg$$20 m g$ of medicine is absorbed.
It is modeled as $x\left(t\right)=100\left(1-0.2t\right)$$x \left(t\right) = 100 \left(1 - 0.2 t\right)$,
that is if
$t=0$$t = 0$, $x\left(0\right)=100mg$$x \left(0\right) = 100 m g$,
$x\left(1\right)=80mg$$x \left(1\right) = 80 m g$,
$x\left(2\right)=60mg$$x \left(2\right) = 60 m g$,
$x\left(2.1\right)=58mg$$x \left(2.1\right) = 58 m g$

In the second case, only part of the medicine comes in contact with the cells and that is absorbed. If the amount of medicine is more, then the amount absorbed is more. As the amount reduces, proportionally the amount absorbed is lesser, because the medicine do not come in contact.

In $1$$1$ hour $100mg$$100 m g$ is reduced to $\frac{100}{1.25}=80mg$$\frac{100}{1.25} = 80 m g$.
And in next 1 hour, it is reduced to $\frac{80}{1.25}=64mg$$\frac{80}{1.25} = 64 m g$

$x\left(t\right)={x}_{0}\frac{1}{a×a×a}={x}_{0}×{a}^{-n}$$x \left(t\right) = {x}_{0} \frac{1}{a \times a \times a} = {x}_{0} \times {a}^{- n}$

This absorption is explained with intervals of 1 hour. But, it does not happen in discrete 1 hour durations, the absorption happens continuously over time. And it is modeled as
$x\left(t\right)={x}_{0}{a}^{-t}$$x \left(t\right) = {x}_{0} {a}^{- t}$

first order

We derived a model for first-order-absorption of medicine.
$x\left(t\right)={x}_{0}{a}^{-t}$$x \left(t\right) = {x}_{0} {a}^{- t}$

This model applies to several other applications.

•  the decay of radio-active material

•  the growth of bacteria when no constraints on resources

•  the compound interest on money charged continuously (not at discrete 3 months or 1 year durations).

These are modeled as
$x\left(t\right)={x}_{0}{a}^{-t}$$x \left(t\right) = {x}_{0} {a}^{- t}$ or
$x\left(t\right)={x}_{0}{a}^{t}$$x \left(t\right) = {x}_{0} {a}^{t}$

Let us examine this further.

rate of change

We derived the model $x\left(t\right)={x}_{0}{a}^{-t}$$x \left(t\right) = {x}_{0} {a}^{- t}$.

The equation takes different values for $a$$a$ for different medicines. For one medicine, $a=10$$a = 10$ and another $a=15$$a = 15$.

It is noted that any number $a$$a$ can be written as $a={b}^{k}$$a = {b}^{k}$, where $b$$b$ does not change and $k$$k$ changes for different $a$$a$.
eg :$10={2}^{3.322}$$10 = {2}^{3.322}$ and $15={2}^{3.907}$$15 = {2}^{3.907}$.

Thus, we are set to define a standard form for $x\left(t\right)={x}_{0}{b}^{-kt}$$x \left(t\right) = {x}_{0} {b}^{- k t}$

Obviously, in the model of ${a}^{-t}$${a}^{- t}$, the variable $a$$a$ varies. But in the model of ${b}^{-kt}$${b}^{- k t}$, the variable $k$$k$ varies. It does not seem to change anything. A value for $b$$b$ is chosen keeping in mind the rate-of-change calculations done in these kind of problems.

Consider $k=1$$k = 1$ and so, $x\left(t\right)={b}^{-t}$$x \left(t\right) = {b}^{- t}$. When the value of $b$$b$ is $2$$2$ or $2.2$$2.2$, the magnitude of rate of change is lower than the $x\left(t\right)$$x \left(t\right)$.
and when the value of $b$$b$ is $3$$3$ or more, the magnitude of rate of change is higher than the $x\left(t\right)$$x \left(t\right)$
There is a value between $2$$2$ and $3$$3$, $e$$e$.

For that value $b=e$$b = e$ in ${b}^{-t}$${b}^{- t}$, the magnitude of rate-of-change of the quantity is proportional to the quantity itself. That happens only when $e=2.71828\cdots$$e = 2.71828 \cdots$.

The value of $e$$e$ is numerically computed, such that rate of change of ${e}^{t}$${e}^{t}$ equals ${e}^{t}$${e}^{t}$.

It is chosen, especially because, it helps in finding rate-of-change(differentiation) and in finding aggregate-of-change (integration). These are explained in later classes.

natural-e

If the mathematical model is
$x\left(t\right)={x}_{0}{a}^{-t}$$x \left(t\right) = {x}_{0} {a}^{- t}$

Then the rate of change is
$\frac{dx\left(t\right)}{dt}=-{x}_{0}×\left(\text{constant}\right)×{a}^{-t}$$\frac{\mathrm{dx} \left(t\right)}{\mathrm{dt}} = - {x}_{0} \times \left(\textrm{c o n s \tan t}\right) \times {a}^{- t}$

The constant is derived to be ${\mathrm{log}}_{e}\left(a\right)$${\log}_{e} \left(a\right)$

If the $a$$a$ is represented with $a={e}^{k}$$a = {e}^{k}$
the equation becomes
$x\left(t\right)={x}_{0}{e}^{-kt}$$x \left(t\right) = {x}_{0} {e}^{- k t}$
$\frac{dx\left(t\right)}{dt}=-{x}_{0}×k×{e}^{-kt}$$\frac{\mathrm{dx} \left(t\right)}{\mathrm{dt}} = - {x}_{0} \times k \times {e}^{- k t}$

The difference between representing the model with $a$$a$ and with $e$$e$ is that,

•  when $a$$a$ is used as the base, the rate of change involves ${\mathrm{log}}_{e}\left(a\right)$${\log}_{e} \left(a\right)$, leading to base of $e$$e$

•  when $e$$e$ is used as the base, the rate of change is readily in the constant $k$$k$.

The latter is considered to be clean.

natural ln

Natural Logarithm : Logarithm of base $e$$e$ is called the natural logarithm.
The natural logarithm is specified as $\mathrm{ln}x$$\ln x$.

examples

The common logarithm of $1000$$1000$ is "$3$$3$".

Common logarithm is to the base $10$$10$.
${\mathrm{log}}_{10}1000=3$${\log}_{10} 1000 = 3$

If natural logarithm of $10$$10$ is $2.3$$2.3$, What is the natural logarithm of $100$$100$?
The answer is "$4.6$$4.6$".

Natural logarithm is to the base $e$$e$.
$\mathrm{ln}100$$\ln 100$
$={\mathrm{ln}10}^{2}$$= \ln {10}^{2}$
$=2\mathrm{ln}10$$= 2 \ln 10$
given that $\mathrm{ln}10=2.3$$\ln 10 = 2.3$
$=2×2.3$$= 2 \times 2.3$
$=4.6$$= 4.6$

summary

Common Logarithm : Logarithm of base $10$$10$ is called the common logarithm.
If the base is not specified then the $\mathrm{log}x$$\log x$ is taken to be common logarithm ${\mathrm{log}}_{10}x$${\log}_{10} x$.

Natural Logarithm : Logarithm of base $e$$e$ is called the natural logarithm.
The natural logarithm is specified as $\mathrm{ln}x$$\ln x$.

Outline

The outline of material to learn "Exponents" is as follows. Note: click here for detailed outline of Exponents s

→   Representation of Exponents

→   Inverse of exponent : root

→   Inverse of exponent : Logarithm

→   Common and Natural Logarithms

→   Exponents Arithmetics

→   Logarithm Arithmetics

→   Formulas

→   Numerical Expressions

→   PEMA / BOMA

→   Squares and Square roots

→   Cubes and Cube roots