PART 1

a) Write a history on logarithm.

History of Logarithms

Logarithms were invented independently by John Napier, a Scotsman, and by Joost Burgi, a Swiss. Napier's logarithms were published in 1614; Burgi's logarithms were published in 1620. The objective of both men was to simplify mathematical calculations. This approach originally arose out of a desire to simplify multiplication and division to the level of addition and subtraction. Of course, in this era of the cheap hand calculator, this is not necessary anymore but it still serves as a useful way to introduce logarithms. Napier's approach was algebraic and Burgi's approach was geometric. The invention of the common system of logarithms is due to the combined effort of Napier and Henry Biggs in 1624. Natural logarithms first arose as more or less accidental variations of Napier's original logarithms. Their real significance was not recognized until later. The earliest natural logarithms occur in 1618.

It can’t be said too often: a logarithm is nothing more than an exponent. The basic concept of logarithms can be expressed as a shortcut……..Multiplication is a shortcut for Addition: 3 x 5 means 5 + 5 + 5Exponents are a shortcut for Multiplication: 4^3 means 4 x 4 x 4Logarithms are a shortcut for Exponents: 10^2 = 100.

The present definition of the logarithm is the exponent or power to which a stated number, called the base, is raised to yield a specific number.The logarithm of 100 to the base 10 is 2. This is written: log10 (100) = 2. Before pocket calculators — only three decades ago, but in “student years” that’s the age of dinosaurs — the answer was simple. You needed logs to compute most powers and roots with fair accuracy; even multiplying and dividing most numbers were easier with logs. Every decent algebra books had pages and pages of log tables at the back. The invention of logs in the early 1600s fueled the scientific revolution. Back then scientists, astronomers especially, used to spend huge amounts of time crunching numbers on paper. By cutting the time they spent doing arithmetic, logarithms effectively gave them a longer productive life. The slide rule, once almost a cartoon trademark of a scientist, was nothing more than a device built for doing various computations quickly, using logarithms.

b) Find and explain the applications of logarithm in two different fields of study.Explanation of each application should include the following

I. The field of study chosen.II. Examples of problem solving related to the field of study

Application of Logarithms

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1. Computational complexity

Analysis of algorithms is a branch of computer science that studies the performance of algorithms (computer programs solving a certain problem). Logarithms are valuable for describing algorithms that divide a problem into smaller ones, and join the solutions of the subproblems.

For example, to find a number in a sorted list, the binary search algorithm checks the middle entry and proceeds with the half before or after the middle entry if the number is still not found. This algorithm requires, on average, log2(N) comparisons, where N is the list's length. Similarly, the merge sort algorithm sorts an unsorted list by dividing the list into halves and sorting these first before merging the results. Merge sort algorithms typically require a time approximately proportional to N · log(N). The base of the logarithm is not specified here, because the result only changes by a constant factor when another base is used. A constant factor, is usually disregarded in the analysis of algorithms under the standard uniform cost model.

A function f(x) is said to grow logarithmically if f(x) is (exactly or approximately) proportional to the logarithm of x. (Biological descriptions of organism growth, however, use this term for an exponential function.) For example, any natural number N can be represented inbinary form in no more than log2(N) + 1 bits. In other words, the amount of memory needed to store N grows logarithmically with N.

2. Probability theory and statistics

Three probability density functions (PDF) of random variables with log-normal distributions.

The location parameter μ, which is zero for all three of the PDFs shown, is the mean of the

logarithm of the random variable, not the mean of the variable itself.

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Distribution of first digits (in %, red bars) in the population of the 237 countries of the world.

Black dots indicate the distribution predicted by Benford's law.

Logarithms arise in probability theory: the law of large numbers dictates that, for a fair coin, as the number of coin-tosses increases to infinity, the observed proportion of heads approaches one-half. The fluctuations of this proportion about one-half are described by the law of the iterated logarithm.

Logarithms also occur in log-normal distributions. When the logarithm of a random variable has a normal distribution, the variable is said to have a log-normal distribution. Log-normal distributions are encountered in many fields, wherever a variable is formed as the product of many independent positive random variables, for example in the study of turbulence.

Logarithms are used for maximum-likelihood estimation of parametric statistical models. For such a model, the likelihood functiondepends on at least one parameter that must be estimated. A maximum of the likelihood function occurs at the same parameter-value as a maximum of the logarithm of the likelihood (the "log likelihood"), because the logarithm is an increasing function. The log-likelihood is easier to maximize, especially for the multiplied likelihoods for independent random variables.

Benford's law describes the occurrence of digits in many data sets, such as heights of buildings. According to Benford's law, the probability that the first decimal-digit of an item in the data sample is d (from 1 to 9) equals log10(d + 1) − log10(d), regardless of the unit of measurement. Thus, about 30% of the data can be expected to have 1 as first digit, 18% start with 2, etc. Auditors examine deviations from Benford's law to detect fraudulent accounting.

PART 2

The volume, V, in cm3, of a solid sphere and its diameter, D, in cm, are related by the equation V=m Dn, where m and n are constants.

Find the value of m and n by conducting the activities below.

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I. Choose 6 different spheres with diameters between 1cm to 8cm. The diameter of the 6 spheres using a pair of vernier calipers.

II. Find the volume of each sphere using water displacement method.III. Tabulate the values of diameter, D, in cm and its corresponding volume, V, cm3.

Find the volume of sphere using water displacement method.

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A method of finding the volume of a sphere with minimal calculations is to use the Water Displacement Method:

1. Fill a beaker or graduated cylinder with enough water to completely immerse the sphere in.

2. Record the baseline initial measurement3. Drop the sphere in4. Record final measurement5. Subtract the initial volume from the final volume ~ this is the volume of the sphere!

1 ml=1 cm3

Value of diameter,D and Volume

Diameter,D (cm ¿ Volume, V (cm3 ¿D1 = 1.0 V1= 0.5

D2 =2.2 V2= 5.5

D3 =3.4 V3= 21.0

D4 =4.8 V4= 58.0

D5 =6.6 V5= 150.0

D6 =7.6 V6= 230.0

Diameter,D (cm ¿ Volume, V (cm3 ¿D1 = 1.0 V1= 0.5

D2 =2.2 V2= 5.5

D3 =3.4 V3= 21.0

D4 =4.8 V4= 58.0

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D5 =6.6 V5= 150.0

D6 =7.6 V6= 230.0

We can solve by simultaneous method

Substitute the values in the equation

V=m Dn

We obtain,

0.5=m¿ ----------(1)5.5=m¿ ----------(2)

m= 0.5

(1.0)n -----------(3)

Substitute (3) into (2)

5.5= 0.5

(1.0)n¿

5.5=0.5¿ 0.5¿

¿

(2.2)n=11

nlog 2.2=log11

n= log 11log2.2

n=3.04 -----------(4)

Substitute (4) into (3)

m= 0.5

(1.0)3.04

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D1 = 1.0 V1= 0.5

D2 =2.2 V2= 5.5

m=0.5

Therefore, m=0.5 and n=3.04

PART 3

3(A)

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D v

1 0.5

2.8 5.5

3.4 21

4.8 58

6.6 150

7.6 230

0 1 2 3 4 5 6 7 80

50

100

150

200

250

f(x) = 0.506266684245238 x^3.02150879657113

Diameter, D

Volu

me,

V

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3(B)

V=m Dn

logV =nlogD+logm

Y=MX +C

Y=logV

X=logD

M=n

C=logm

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0.5

0

0.5

1

1.5

2

2.5

3

f(x) = 3.02150879657112 x − 0.295620651172346

Diameter, D

log

V

3c) From the graph, find

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log D log V0 -0.30103

0.447158 0.7403630.531479 1.3222190.681241 1.7634280.819544 2.1760910.880814 2.361728

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0.5

0

0.5

1

1.5

2

2.5

3

f(x) = 3.02150879657112 x − 0.295620651172346

Diameter, D

log

V

1. The value of m and of n, thus express V in terms of D.

V=m Dn logV =nlogD+logm

n is the gradient of the graph

n=2.176091−1.3222190.819544−0.531479

n=0.8540.288

n=2.964 ≈ 3.0 (nearest whole number)log m isthe y−intercept of the graph log m=−0.30

m=10−0.3=0.4999999 ≈ 0.5

V=0.5 D 3

2. Volume of the sphere when diameter is 5cm

Since graph is logV against logD, we need to transfer, D=5cm int0 logD=log5=0.6989

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log D log V0 -0.30103

0.447158 0.7403630.531479 1.3222190.681241 1.7634280.819544 2.1760910.880814 2.361728

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0.5

0

0.5

1

1.5

2

2.5

3

f(x) = 3.02150879657112 x − 0.295620651172346

Diameter, D

log

V

We get log v=1.79

V=101.79 .

Volumeof sphere=61.65 cm3

3. The radius of the sphere when the volume is 180 cm3

Change to logv=log180=2.25, From the graph, we get log D=0.845

D=100.845

D=6.99 2 r=6.99 D=2r

r=6.99/2r=3.499cm

FURTHER EXPLORATION

a) V=0.5 D 3 -------(1) V= 43

π r3 ------------(2)

(1)=(2)

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0.5 D3 ¿43

π r 3 D¿2 r

0.5(2r )3 ¿43

π r 3

0.5(8)r3 ¿43

π r 3

4 r3 ¿43

π r 3 -------------------cancel r3 on both sides

4 ¿43

π

π=4( 34)

π=3

b) Another method to find value of πis using Monte Carlo simulation or Archimedes method of Exhaustion

REFLECTION

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Symbols used in this project using Microsoft word equation insert tool really help me so much here are some of the symbol I use.

V= 43

π r3

logV =nlogD+logm

0 1 2 3 4 5 6 7 8 90

50

100

150

200

250f(x) = 0.510817713781475 x^2.98842704058272

Diameter, D

Volu

me,

V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.5

0

0.5

1

1.5

2

2.5

3

f(x) = 2.98842704058272 x − 0.291734050985445

Diameter, D

log

V

I really learn how to use Microsoft excel and word to do graph, insert equation and a lot more.

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