Appendix 1:
Understanding Logarithm

Many processes are best
described by logarithmic based equations and it is essential that students have
an understanding of logarithmic scales and numbers. There are two common logarithms in use and
both can be accessed through a scientific calculator. The decimal base 10 scale, represented by the
"log" and 10^{X} keys on your calculator and the natural
logarithm or base e scale, which is represented by the "ln" and e^{x}
on your calculator. Although calculus is
not a prerequisite to this class, you will be required to memorize many
logarithmic relationships that have been derived from calculus. Thus it is imperative that you know the basic
algebraic operations required to manipulate logs. Let’s start from the
beginning.

Addition takes two or more
numbers and adds their values to form a third.

2+2=4

Multiplication is the
repetition of addition

2x4 = 2+2+2+2 = 8

2xn repeats
the process of adding two n times.

Division represents the
reciprocal of multiplication, taking a number and dividing by 2 determines how
many times you have to add two to get that number.

8/2=4 so 2x4=8 or 2+2+2+2=8

note; taking any number (Z) and
multiplying and dividing by any other equivalent number (n), gives back that
number back, so they are inverse operations:

(Z x n)/n=Z

Exponentiation represents the
repetition of multiplication as multiplication represents the repetition of
addition.

2x2x2x2x2x2 = 2^{6} = 64

X=b^{n} is the process of multiplying b by
itself n times

Logarithms represent the
inverse of exponentiaion.

log_{2}64=6 so 2^{6}=64=2x2x2x2x2x2

log_{b}(X)=n

where b is a real number called the base of
the log (the number being multiplied by itself) n times to get X. Exponentiation
is often called the anti log.

If Y=log_{b} X

then X is the antilog_{b}
of Y,

or X=b^{Y}

As this is the inverse of exponentiation

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x = log10^{x} = ln(e^{x})

**Mathematical manipulations**

Logarithms reduce the process
of multiplication and division to those of addition and subtraction (we
add/subtract logs like we multiply/divide their numbers). Conisder

4x8=?

From above we would determine
this is 2^{2}x2^{3}.
Look at the exponents, this is

2^{2+3}=2^{5}=32

So we added exponents to
perform multiplication.

Likewise, we could have taken
the log_{2 }of the two numbers and added them, and then taken the
antilog of the sum

log_{2}(4) = 2, log_{2}(8) = 3

2+3=5, the antilog (base 2) of 5 is: 2^{5}=32

You are required to know
these operations.

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For example if a^{m}a^{n}
= a^{(m+n)} makes no sense to you, try it with base 10.

10^{2}x10^{3}
= 10^{(2+3)} = 10^{5} =
100,000 (ie., a hundred times a thousand is a hundred thousand)

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**Comparison of Logarithms to Scientific Notation:** Let’s consider
two numbers; 0.0003547 and 46794.

In both scientific notation
and logarithms any number is be broken into 2 parts.

In scientific notation these
are 10 to a power and a number between 1 and 10. That is Ax10^{X}_{ } where 1<A<10. The value of A can be expressed as 10 to the
power of a fraction (M –the mantissa of a log)

For logarithms these are the
characteristic - C (an integer) and the mantissa –M (a fraction, whose antilog
is a number between 1 and 10).

We write this as C.M or the characteristic.mantissa

In scientific notation 467940=
4.6794x10^{5} =10^{M}10^{X} = 10^{.67019}10^{5}

For numbers larger than 10,
X=C, the characteristic of the log and in logarithms this is C.M or 5.67019

so 467940= 4.6794x10^{5}
=10^{5.67019} and the log of 467940 is 5.67019

where the characteristic=5
and the mantissa=0.67019

To take the antilog of C.M,
you have

10^{C.M}=10^{C}10^{M}=10^{M}10^{C}
= a number between 1and 10 x 10^{X}

Example, the antilog of 5.67019
= 10^{5.67019 }= (10^{5} )(10^{.67019} )=10^{.67019}x10^{5
}= 4.6794x10^{5}

For fractions (negative logs)
things are a bit more confusing. The trick is to realize the mantissa can never
be negative as 10 to the power of the mantissa represents a number between 1
and 10.

Consider the fraction 0.00003457.

Now first, 0.00003457 =
3.457x10^{-5 }.

Now, this is greater than 10^{-5}
and smaller than 10^{-4}, and so -4 and not -5 is the characteristic. To see this we can break it scientific
notation, express them as powers of 10, and then add the exponents.

0.00003457 = 3.457x10^{-5 } = 10^{M}10^{X} =10^{.5387}(10^{-5
})=10^{.5387-5}=10^{-4.4613}

So the log of 0.00003457 = -4.4613
and

0.00003457 = 3.457x10^{-5 } = 10^{-4.4613}

Now note, the antilog of -4.4613=
(10^{-4.4613 })= 10^{M}10^{C} = 10^{M+C} where
M must be a positive fraction. So C = -5 and -3.4613=-5+M and M = .5387, giving

(10^{-4.4613 })= 10^{-5+.5387}=)=
10^{-5}10^{.5387}= 10^{.5387}10^{-5} =3.457x10^{-5}

Lets apply the mathematical
arules to the two state Arrhenius equation which relates the kinetic rate
constant to the temperature. If a
reaction has an activation energy of 40 kJ/mol at room temperature (20^{o}C),
at what temperature would the rate double if all other variables were kept
constant. That is, at what temperature
(T_{2}) does k_{2}= 2k_{1}

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Dividing the first state by k_{2:}

_{ }

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Substituting for k_{2 }

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Taking the natural log of
both sides:

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Derivation of Beer's Law: (You are not required to be able to do this,
but this shows why this class uses so many logarithms).

Beers law is based on a few
basic assumptions and its derivation requires calculus like many of the
equations we will using this semester.
Calculus is not a prerequisite for this class and you can memorize
beer's law. To understand beer's law we
ask ourselves what variables influence the amount of light absorbed by a solution. There are 3 variables for a given wavelength;
the concentration (C) of absorbing compounds (chromophores), the distance the
light travels (X) in the solution (path length) and the intensity (I)
of light itself (noting that intensity of light is related to the number of
photons while the energy is related to the wavelength). Mathematically we use a proportionality
constant (k) to state that the change in light intensity is proportional to
these 3 factors.

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The negative sign is a result
of the fact that the intensity decreases as the light is absorbed.

The solution of this equation
requires the mathematics of change, calculus, where dI represents an
infinitesimally small change in the intensity correlated with an infinitesimally
small change in the path length (dx):
Once again we see why logarithms are so important in science. Note that in going from the natural logs of
the calculus to the log base 10 of beers law we use the relationship
lnX=2.303log x and so k = 2.303a.

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