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 = 2⁶ = 64

X=bⁿ is the process of multiplying b by itself n times

Logarithms represent the inverse of exponentiaion.

log₂64=6 so 2⁶=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

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 2²x2³. Look at the exponents, this is

2²⁺³=2⁵=32

So we added exponents to perform multiplication.

Likewise, we could have taken the log₂ of the two numbers and added them, and then taken the antilog of the sum

log₂(4) = 2, log₂(8) = 3

2+3=5, the antilog (base 2) of 5 is: 2⁵=32

You are required to know these operations.

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

10²x10³ = 10⁽²⁺³⁾ = 10⁵ = 100,000 (ie., a hundred times a thousand is a hundred thousand)

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= 4.6794x10⁵ =10^M10^X = 10^.6701910⁵

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

so 467940= 4.6794x10⁵ =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^C10^M=10^M10^C = a number between 1and 10 x 10^X

Example, the antilog of 5.67019 = 10^5.67019 = (10⁵ )(10^.67019 )=10^.67019x10⁵ = 4.6794x10⁵

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^M10^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^M10^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^-510^.5387= 10^.538710^-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^oC), at what temperature would the rate double if all other variables were kept constant. That is, at what temperature (T₂) does k₂= 2k₁

Dividing the first state by k_2:

Substituting for k₂

Taking the natural log of both sides:

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.

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.