This document provides definitions and examples of linear equations, quadratic equations, and their applications in business mathematics. Linear equations involve variables raised to the first power and are used to model relationships that vary proportionally. The document defines slope, linear functions, and using linear equations to analyze break-even points. Quadratic equations involve variables raised to the second power and are used to model nonlinear relationships. Examples are provided of solving quadratic equations through factoring and applications to profit modeling in business.

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Definition of Business Mathematics:
Business mathematics is the branch of mathematics which is helpful for businessmen. Every
businessman is in tension about how to solve different business problem. In these problems we
can include , interest calculation , hire rates , salary calculation, tax calculation , Public provident
fund calculation , Foreign exchange calculation and more over what is the quantity he should
produce etc. Business mathematics is not only helpful for businessmen but also it is helpful for
general public because, if they learn it, they can deal with businessmen with better way. Business
mathematics is easier than general mathematics. It can learn businessmen very easily. At
Graduation and post graduation level, it is the main subject of commerce .Business's all part like
Man, money machine and material depends on business mathematics. There are following
feature of business mathematics:-
1. It is totally related to business and solution of business problem.
2. It is main branch of general mathematics.
3. Statistics also include in it.
4. Its main feature is also that variety of formulae are in this mathematics , so businessmen
have higher option to choose any one best formula for solve their business problem
“Business mathematics is all about how the problems from the realm of business and
technology can be translated into the language of mathematics and then be solved by means of
mathematical models. Quite often this does not involve only one single mathematical model -
what makes the matter even more exciting’’.
“Business math’s are mathematics used by commercial enterprises to record and
manage business operations. Commercial organizations use mathematics in accounting,
inventory management, marketing, sales forecasting, and financial analysis. Mathematics
typically used in commerce includes elementary arithmetic, elementary algebra, statistics and
probability. Business management can be made more effective in some cases by use of more
advanced mathematics such as calculus, matrix algebra and linear programming”.

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Number System
A number is a mathematical object used to count, label, and measure. In mathematics, the
definition of number has been extended over the years to include such numbers as 0, negative
numbers, rational numbers, irrational numbers, and complex numbers.
Mathematical operations are certain procedures that take one or more numbers as input and
produce a number as output. Unary operations take a single input number and produce a single
output number. For example, the successor operation adds 1 to an integer, thus the successor of 4
is 5. Binary operations take two input numbers and produce a single output number. Examples of
binary operations include addition, subtraction, multiplication, division, and exponentiation. The
study of numerical operations is called arithmetic.
Important number systems:
Natural 0, 1, 2, 3, 4, ... or 1, 2, 3, 4, ...
Integers ..., −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, ...
Rational a/b where a and b are integers and b is not 0
Real The limit of a convergent sequence of
rational numbers
Complex a + bi or a + ib where a and b are real
numbers and i is the square root of −1
Importance of Number System:
Numbers are so fundamental that we are using them every time in the form of units of
measurements of mass, space and time. It will at time explained by symbols only. We initiate the
number system by natural numbers and then proceed on to other system.

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Progression
Example of Arithmetic progression:
Mr. Anwar Hossain works as an assistant secretary in Prime Islami Life Insurance Company. He
joined the company in 1997 after completing his MBA from Dhaka University. After working
two years he gradually realized to save money for his future life. He discussed elaborately the
matter with his friends who had already started saving.
After getting inspiration from his friends he started saving in Popular Life Insurance Company.
He saved tk.1, 65,000 in ten years. In each year after the first he saved tk.1000 more than he did
in the preceding year. He saved tk.12000 in the first year. To find his saving in the first year he
used the summation formula of Arithmetic Progression.
Problem:
Anwar Hossain saved tk.160, 500 in 10 years. In each year after the first he saved tk.1000 more
than he did in the preceding year. How much did he save in the first year?
Solution:
This problem can be solved by using arithmetic progression. At first we have to define all the
mathematic terms found in the problem.
Here,
n = number of years = 10
[This is the period of his saving i.e. he saved tk.160, 500 within 10 years. This time period may
vary from person to person.]
Again,
d = common difference = 1,000

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That means in each year after the first he saved tk.1000 more than he did in the preceding year.
This amount is constant or same for the long 10 years. If it varies then the law of arithmetic
progression will not work.
Now,
Sn = Sum of money saved = tk. 16, 500
In arithmetic progression sum of amount is calculated by a special formula i.e.
Sn = n/2 [2a + (n-1) d]
Here, the sum of the amount is found out by this formula.
At last, a = saving in the first year =?
This is the amount which he first saved. This amount is to find out. To find out the savings in the
first year we have to use the formula of sum. That is
Sn = n/2 [2a + (n-1) d]
Or, 1, 65,000 = 10/2 [2a + (10-1) 1000]
0r, 165,000 = 5 [2a + 9000]
Or, 165,000 = 10a + 45000
Or, 10a = 165,000-45,000
Or, 10a = 120,000
Or, a = 120,000/10
Or, a = 12,000
Here,
n = number of year

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a = saving in the first year
Sn = summation
d = common difference
Therefore,
a = savings in the first year = 12,000
So Mr. Anwar Hossain saved tk. 12,000 in the first year.
Geometric Progression
Definition
A geometric Progression is a sequence whose term increase or decrease by a constant called
common ratio. A series is geometric progression thus is a multiplicative series whose common
ratio can be found by dividing any term by its preceding term.
Thus the sequence 1, 2, 4, 8, 16, 32…….. is an infinite geometric progression, the first term is 1
and the common ratio is 2.
If for a sequence, un+l remains constant for all natural numbers n, then the sequence is called
G.P.
We can also state the series as,
Sn = a+ar+ar2+ar3+………. +ar n-3+ar n-2+ar n-1
Sum of a series in G.P:
The sum of n terms of a series is G.P. can be found out using the following formula :
Sn = a (rn-1)/r-1, when r>1
Or, Sn =a(1-rn)/1-r,when 1>r

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Example of Geometric Progression
If the value of spinning machine of Padma Poly Cotton Knit Fabrics Ltd. depreciated by 25
percent annually, what will be its estimated value at the end of 8 years if its present value is Tk.
204800 ?
Solution:
Geometric progression can be used to solve the problem. After simplifying the problem partially
we can use geometric progression. At first we have to define all the terms given in the problem.
That will help us to calculate the problem properly.
Here,
Present value of car = Tk. 204800
Present value means the value which is available now i.e. which is not depreciated through use.
Again,
Value of machine deprecated annually =25%
We know, after using a machine or assets it is depreciated day by day. So the Spinning machine
is also depreciated by 25 percent annually.
Here, n = number of years = 8
Now, If the present is 100, then value after one year = Tk. 75
If the present is 1, then value after one year = 75/100
If the present is 204800 then value =Tk. (75/100)*204800
=Tk. 153600
a =1536000

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We also note that values at the end of second, third, fourth, fifth, sixth, seventh and eighth years
from a G.P. with common ratio, = 75/100 = ¾
Common ratio is a fundamental characteristic of G.P. on the other hand common difference is for
A.P.
Now, value at the end of eight year can be found out by using the formula from G.P.
The formula is ar n-1
Therefore, ar n-1
=ar 7
=153600 × ( 3/4)7
= Tk. 20503.125
So, the estimated value will be Tk. 20503.125 at the end of 8 years.

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Linear Equations:
Objective of the study:
Best way CNG Filling Station
By using Linear Equations
ß To determine cost, profit, revenue functions.
ß To determine net profit for Best Way CNG Filling Station before taxes.
ß To analyze Break-Even-Points & Equilibrium Points.
By using Differential Calculus
ß To determine the new area of the factory building for Best Way CNG Filling Station.
ß To test the local minima and to determine new value of length and breadth.

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Linear Equations and Functions:
Linear programming is used to solve problems in many aspects of business administration
including:
ß Product mix planning
ß Distribution networks
ß Truck noting
ß Staff scheduling
ß Financial portfolios
ß Corporate restriction
A common form:
Thus, 2x – 3y = 7 is a linear equation because it consists of the constant 7, the term 2x (which is
the constant 2 times x to the first power), and -3y, which is also a term consisting of a constant
times on variable to the first power.
y= -x+5
y= 0.5x+2

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Slope = Vertical Change
Horizontal Change
Slope = m = Difference of Y’s
= y
2- y1
Difference of x’s x
2
- x
1
Break-even Analysis:
Total Revenue = Total Cost
=> R (Q) = C (Q)
=> Price x Quantity = Total Fixed Cost + (Variable Cost x Quantity)
=> P x Q = TFC + (V x Q)
=> P x Q- (V x Q) =TFC
=> Q (P-V) = TFC
..
. Break-even Quantity= TFC/ (P-V)
∑ k(-5,6)
∑ L(-4,4) 6= Vertical change (rise)
∑ Q(-6,2) * J (-2,2) 2= Horizontal change (run)
° F(1, -1)
∑ H(-3, -3) ° E(1,-3) ° D(7.5-3)
∑ G(-4,-6) ° C(2.5, -7)
Cost/PFCS
(TC, PER UNIT Revenue
VC
BEP
150000 FC
0

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LINEAR EQUATION TO SOLVE BUSINESS PROBLEMS:
Best way CNG Filling Station expects fixed cost of $ 22,800. Margin is to be 55% of retail.
Variable cost in addition to costs of goods is estimated at $ 0.17 per dollar of sales.
Requirements:
So, Cost function = C(x) = 22,800+ 0.17 x + 0.45x
C(x) = 22,800 + 0.62 x
Revenue function = R(x)= x
So, the profit function = R(x)- C(x)
= x –x 22,800 – 0.62x
P(x) = 0.38x – 22,800
Find the revenue
cost, and functions Find the break-even
point
Determine net profit
before taxes be
on sales of 75,000?
Show the
break-even
chart.
Fixed cost for goods =$ 22,800
Variable cost = 0.17x
Margin = 55% of retail
Cost of goods sold = Retail Price – Margin
= S- 0.55x
= 0.45x

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Break-even Point For Best way cng Filling Station:
According to the theory of Break-even we know that at the point of break-even, P (x)=0
Profit for Best Way Cng Filling Station
Break-even Point Chart:
Profit function = P(x) = 0
 0.38x – 22,800 = 0
 x = 60,000
..
. Break-even Quantity = 60,000
So, the break-even dollar volume of sales is,
R(60,000)$ 60,000
So, the break-even point is (60,000, 60000)
The net profit before taxes be on sales of 75,000-22,8000 is
P(75,000) = 0.38 x 75,000-22,800
= $ 5,700
Revenue
and Cost
100,000 - Revenue: R(x)= x
Break-even Point
75,000 - (60,000, 60000
50,000 - Cost: C(x)= 0.62x+ 22,800
0 50,000 100,000 Sales

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Quadratic Equation:
The name Quadratic comes from "quad" meaning square, because the variable
gets squared (like x2).
It is also called an "Equation of Degree 2" (because of the "2" on the x)
Standard Form
The Standard Form of a Quadratic Equation looks like this:
ax2
+bx+c=0
a, b and c are known values. a can't be 0.
"x" is the variable or unknown (we don't know it yet).
Here are some more examples:
2x2
+ 5x + 3 = 0 In this one a=2, b=5 and c=3
x2
− 3x = 0 This one is a little more tricky:
Where is a? Well a=1, and we don't usually write "1x2"
b = -3
And where is c? Well c=0, so is not shown.
5x − 3 = 0 Oops! This one is not a quadratic equation: it is missing x2
(in other words a=0, which means it can't be quadratic)
Hidden Quadratic Equations!
So the "Standard Form" of a Quadratic Equation is
ax2
+ bx + c = 0

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But sometimes a quadratic equation doesn't look like that! For example:
In disguise → In Standard Form a, b and c
x2
= 3x -1 Move all terms to left hand side x2
- 3x + 1 = 0 a=1, b=-3, c=1
2(w2
- 2w) = 5 Expand (undo the brackets),
and move 5 to left 2w2
- 4w - 5 = 0 a=2, b=-4, c=-5
z(z-1) = 3 Expand, and move 3 to left z2
- z - 3 = 0 a=1, b=-1, c=-3
5 + 1/x - 1/x2
= 0 Multiply by x2
5x2
+ x - 1 = 0 a=5, b=1, c=-1
It is important to realize that these equations open up a world of real-life applications especially
those where the equations are usually not "nice" and factorable. Here we present a taste of such
applications.
Solving Quadratic Equations by Factoring:
An application from business. The profit from the sale of x king size Snickers bars at a ball
game is given by P = -0.001x2
+1.6x-100. What level(s) of sales yield a profit of $500?
We want to find x where P=500. So we need to solve 500 = -.001x2
+1.6x-100.
500 = -.001x2
+ 1.6x - 100
Move everything to one side: -.001x2
+ 1.6x - 100 - 500 = 0
Simplify: -.001x2
+ 1.6x - 600 = 0
Mult thru by -1000 to clear decimals: -1000[-.001x2
+ 1.6x - 600] = -1000[0]
Distribute: x2
- 1600x + 600,000 = 0
Factor: (x - 600)(x - 1000) = 0
and Solve: x = 600, 1000
So selling either 600 or 1000 Snickers bars will yield a profit of $500. Here are some things to
think about and discuss on the board if you would like. Why is it better to multiply through by -
1000 instead of 1000? How could the profit be the same from two different levels of sales? Why

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wouldn't it be that the more bars you sell the greater the profit? What happens to profit if you sell
more than 1000 bars? How could you find the level of sales needed to maximize profit? (This
last question we will answer in a later lesson!)
Solving Quadratic Equations by the Square Root Method:
An application from meteorology. Weather forecasters frequently report wind chill factors in
addition to air temperatures. A body exposed to wind loses
heat due to convection. The amount of loss depends on many factors, one of them is a positive
number called the coefficient of convection, Kc. The relationship between Kc and wind velocity
v (in mph) is given approximately by (Kc)2
/64 - 1/4 = v. As the wind velocity increases from 10
mph to 40 mph, by approximately how much does Kc increase?
Let's solve for Kc when v=10 and when v=40 and compare our results. Note the use of the
square root method of solving quadratic equations.
(Kc)2
/64 - 1/4 = 10 (Kc)2
/64 - 1/4 = 40
64[(Kc)2
/64 - 1/4] = 64[10] 64[(Kc)2
/64 - 1/4] = 64[40]
(Kc)2
- 16 = 640 (Kc)2
- 16 = 2560
(Kc)2
= 656 (Kc)2
= 2576
Kc = ± sqrt(656) ± 25.6 Kc = ± sqrt(2576) ± 50.8
We can ignore the negative answers since we were told that Kc is a positive number. So, as the
wind velocity increases fourfold from 10 to 40 mph, the coefficient of convection approximately
doubles. [This example is based on Harshbarger/Reynolds' Mathematical Applications.

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Solving Quadratic Equations by Completing the Square:
An application from economics. A demand equation gives the relationship between the price p
of a good and the quantity q that a consumer is willing to purchase in a fixed period of time. The
law of demand says that as the price increases, the quantity demanded by the consumer falls
(makes sense), and visa versa. A supply equation gives the relationship between the price p of a
good and the quantity q that a manufacturer is willing to produce. The law of supply says that as
the price rises, the quantity the producer is willing to supply rises, and visa versa. If the demand
equation for a commodity is p(q+4)=400 and the supply equation is 2p-q=38, find the market
equilibrium.
At market equilibrium, both equations will have the same p-value. So lets solve the supply
equation for p and substitute it into the demand equation. Then we will have a single-variable
equation quadratic in q.
2p - q = 38 2p = q + 38 ½[2p] = ½[q + 38] p = ½q + 19
Plugging that in to p(q+4)=400 for p gives:
(½q + 19)(q + 4) = 400 q2
+ 42q= 648
½q2
+ 2q + 19q + 76 = 400 q2
+ 42q + 441 = 648 + 441
½q2
+ 21q - 324 = 0 (q + 21)2
= 1089
2[ ½q2
+ 21q - 324] = 2[0] q + 21 = ± sqrt(1089)
q2
+ 42q - 648 = 0 q + 21 = ± 33
This may factor, but 648? Yuck! q = -21 ± 33
Let's complete the square. q = 12, -54

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A negative amount of goods is silly, so the only relevant answer is 12. What price is associated
with q=12? Plug it back into p=½q+19 which gives p=½(12)+19=25. So, market equilibrium
occurs when 12 units are sold at a price of $25 each.
Let's check our algebraic analysis by looking at the graphs of the supply and demand functions.
Let the vertical axis be p and the horizontal axis be q. Graph the supply function p=½q+19 and
the demand function p=400/(q+4). Use your graphing calculator to find the intersection point.
Looks good!
Think about why the graph of the demand function is decreasing and the supply function is
increasing. Is this always true? [This example is based on Harshbarger/Reynolds' Mathematical
Applications.
The completing the square method is useful at times, but the quadratic formula is more often
the method of choice.

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Solving Quadratic Equations using the Quadratic Formula:
An application from biology. The metabolic rate of ectothermic organisms increases as the
temperature increases within a certain range. The following data gives the oxygen consumption
of a beetle at certain air temperatures.
Temperature
(degrees Celsius)
Oxygen Consumption
(Microliters per gram per hour)
10 80
15 127
20 198
25 290
These data can be modeled by the quadratic function C=0.45T2
-1.65T+50.75 for 10
T 25 where T is temperature and C is
consumption. Find the air temperature (to the nearest tenth of a degree) when the beetle's
oxygen consumption is 150 microliters per gram per hour.
We want to find T where C=150. That is, we need to solve the quadratic equation
150=0.45T2
-1.65T+50.75.
a=9, b=-33, c=-1985
By the quadratic formula,
150= 0.45T2
- 1.65T + 50.75
0.45T2
- 1.65T - 99.25 -150=0
.45T2
-1.65T – 99.25=0
100[0.45T2
- 1.65T - 99.25] = 100[0]
45T2
- 165T - 9925 = 0
[45T2
- 165T - 9925]/5 = 0/5
9T2
- 33T - 1985 = 0

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-13.1o
is out of the range for this function (which was 10 T 25) so we ignore it and the final
answer is 16.8o
C. Let's check our result by graphing the function and analyzing the graph at
C=150. Graph y1=0.45x2
-1.65x+50.75 on your calculator and TRACE to the point where y=150
(or as close as you can get). Is the x-value (approximately) 16.8? Woohoo!
Definition of SET:
A set is a group or collection of objects or numbers, considered as an entity unto itself. Sets are
usually symbolized by uppercase, italicized, boldface letters such as A, B, S, or Z. Each object or
number in a set is called a member or element of the set. Examples include the set of all
computers in the world, the set of all apples on a tree, and the set of all irrational numbers
between 0 and 1.
When the elements of a set can be listed or denumerated, it is customary to enclose the list in
curly brackets. Thus, for example, we might speak of the set (call it K) of all natural numbers
between, and including, 5 and 10 as:
K = {5, 6, 7, 8, 9, 10}
A set can have any non-negative quantity of elements, ranging from none (the empty set or null
set) to infinitely many. The number of elements in a set is called the cardinality, and can range
from zero to denumerable infinite (for the sets of natural numbers, integers, or rational numbers)

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to non-denumerable infinite for the sets of irrational numbers, real numbers, imaginary numbers,
or complex numbers).
The Importance of Set Theory
One striking feature of humans is their inherent need –and ability –to group objects according to
specific criteria. Our prehistoric ancestors grouped tools based on their hunting needs. They
eventually evolved into strict hierarchical societies where a person belonged to one class and not
another. Many of us today like to sort our clothes at house, or group the songs on our computer
into playlists.
The idea of sorting out certain objects into similar groupings, or sets, is the most fundamental
concept in modern mathematics. The theory of sets has, in fact, been the unifying framework for
all mathematics since the German mathematician Georg Cantor formulated it in the 1870’s.No
field of mathematics could be described nowadays without referring to some kind of abstract set.
A geometer, for example, may study the set of parabolic curves in three dimensions or the set of
spheres in a variety of different spaces. An algebraist may work with a set of equations or a set of
matrices. A statistician typically works with large sets of raw data. And the list goes on. You
may have also read or heard that the most important unresolved problem in mathematic sat the
moment deals with the set of prime numbers (this problem in number theory is known as
Riemann’s Hypothesis; the Clay Institute will award a million dollars to any person that solves
it.) As it turns out, even numbers are described by mathematicians in terms of sets!
More broadly, the concept of set membership, which lies at the heart of set theory, explains how
statements with nouns and predicates are formulated in our language–or any abstract language
like mathematics. Because of this, set theory is intimately connected to logic and serves as the
foundation for all of mathematics.

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Example:
Out of 500 customers of a locality, 200 like Olympics pen, 150 like battery and
120 like confectionary products, of the total 100 like both pen and battery, 80 like
both confectionary and battery and 50 like both confectionary and pen. 40 like all
of the three products.
Let, P.B and C denote the set of customers like Olympics pen, battery and
confectionary products & S represents the total number of customer.
So,
n (P)= 200; n(B)= 150; n (C)= 120; n(S)= 500
n (P«B)= 100; n(B«C)= 80; n(P«C)= 50
n (P«B«C)=40
So, number of customers who like at least one product is given by
n (P»B»C)=n(P)+n(B)+n(C)- n(P«B)- n(B«C)- n(P«C)+ n(P«B«C)
= 200 + 150 + 120 – 100 – 80 - 50+40
= 280
So, the number of customers who did not like any product,
= n (S) - n (P»B»C)
= 500 - 280
= 220

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So, among 500 people, 280 customers want to purchase Olympics’ Products.
Short Findings:
i) We calculate easily of product element.
ii) In the Venn Diagram we easily determine the difficult problem.
iii) The manager can easily separate and mixed the product.
Importance of Mathematics of Finance:
Financial Mathematics and Business Mathematics from two important branches of mathematics
in today's world and these are direct application of mathematics to business. Organizations
routinely use common reporting tools to help with the technical processing of data (e.g., profit
and loss reports, balance sheets, etc.) for the purpose of management accounting. And
management can usually obtain whatever information they need for their internal decision
making.
Financial Mathematics tools can go beyond the scope of such common reporting to elaborate
financial for external users, such as, for example, decision makers, stakeholders, and suppliers.
Financial mathematics are tools used in the valuation and the determination of yields on
investments and costs of financing arrangements. In this chapter, we introduce the mathematical
process of translating a value today into a value at some future point in time, and then show how
this process can be reversed to determine the value today of some future amount. We then show
how to extend the time value of money mathematics to include multiple cash flows and the
special cases of annuities and loan amortization. We then show how these mathematics can be
used to calculate the yield on an investment.

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The notion that money has a time value is one of the most basic concepts in investment analysis.
Making decisions today regarding future cash flows requires understanding that the value of
money does not remain the same throughout time.

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Example:
Bank of Asia offers 7.6 percent compounded continuously and uses the modified year. Find the
effective rate.
Solution:
Give,
Nominal rate= 7.6% = 0.076
Effective rate, re= ?
For multiplier is= 365÷ 360
..
. Replacing exponet is 365 ÷ 360 x (0.076)= 0.07705555
Instead of 0.076 for the modified
Year Facility
..
. re ej
– 1
=

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Short Findings:
i) We can easily determine the production profit or loss.
ii) We understand actual interest of the loan or borrowed money.
iii) We can easily determine what is the time is required to essential of production.

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Definition of Logarithm:
The logarithm of a positive real number x with respect to base b, a positive real number not equal
to 1[nb 1]
, is the exponent by which b must be raised to yield x. In other words, the logarithm of x
to base b is the solution y to the equation[2]
The logarithm is denoted "logb(x)" (pronounced as "the logarithm of x to base b" or "the base-b
logarithm of x"). In the equation y = logb(x), the value y is the answer to the question "To what
power must b be raised, in order to yield x?". This question can also be addressed (with a richer
answer) for complex numbers, which is done in section "Complex logarithm",
The Importance of Logarithms
Logarithms are a function which relates that were formulated by John Napier in the beginning of
the seventeenth century. His original purpose was to make calculations easier. Since then,
engineers, mathematicians, and scientists have been using them for a variety of purposes. The
slide rule is based on this principle; this used to be a standard instrument for scientists to make
calculations when time was more valuable then precision until more efficient methods became
available. The main reasons that logarithms simplify expressions are due to the product, quotient,
and power rules.
Logarithmic scales show up in sensory perception and other scales. Humans perceive both sound
waves and light waves logarithmically. Additionally, the Richter magnitude scale is logarithmic
base 10 scale with compared to standard 0, while the pH scale denotes the relative amount of
hydrogen ions and describes the acidity or basicity of a solution.
The natural logarithm is the most important logarithm in mathematics due to its property that its
derivative is the inverse function. While also being the inverse of the exponential function. This
is one of the ways the natural logarithm can be defined. One of the interesting features of
logarithms is that they only differ by a factor, due to the change of base rule: $log_{b}x =

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log_{a} x/log_{a} b$. Because of this, an efficient method for computing one logarithm allows
for the efficient computation of all logarithms. The other important logarithms are the common
(base 10) logarithm, due to the use of the decimal system, and the base 2 logarithm, which is
important to computer scientists.
Logarithms, and particularly the natural logarithm, are integral to thermodynamics because it
relates the multiplicity of the system to its entropy, which allows us to rigorously determine the
temperature of the system. Without the tools that logarithms offer the sciences today, the
understanding of the world would be greatly hindered.
Example:
Find the compound interest on Tk. 10,000 for 4 years at 5% per annum. What will be the simple
interest in the above case?
Solution:
Here, P= 10,000, n=4, i= 5/100= 0.05
We know A= P(1+i)n
..
. A = 10,000 (1+.05)n
= 10,000(1.05)4
= 10,000 x 1.215
= 12150
C.I = Compound Interest
= A- P= Tk. 2150
S.I = Simple Interest
= P.n.i= 10,000 x 4 x (5÷100)
= Tk. 2000
x = (1.05)n
Log x = 4 log 1.05
= 4 x 0.0212
= 0.0848
x = anti-log (0.0848
= 1.215

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Short Findings:
i) We can easily determine of mathematics problems.
ii) We can calculate the exact value of math’s.
iii) In the business prospective the exact value is too much important.
Conclusion:
Business Mathematics in BBA student is able to make more effective in some cases by use of
more advanced mathematics such as calculus, matrix algebra and linear programming logarithm
set theory, permutation & combination etc. Business organizations use mathematics in
accounting, inventory management, marketing, sales forecasting, financial analysis etc.
Importance of Mathematics is a scheme of Business Management for BBA student.