Relevance and Application of Exponential Functions in Real Life Situations

In this article relevance and application of exponential functions in real life situations , we will see how exponential function used in real life. Exponential function is used in more real life application like compound interest, problem based on population, problem based on radioactive decay, mortgage problems. Let us work out some problem to make understand that application of an exponential function.

Exponential Function Growth:

Exponential function growth`g=c(p)^t where,`

c-Number at initial

p-growth factor (p =1+r,here r is given growth percentage)

t-time

Example 1 -  Relevance and application of exponential functions in real life situations

A group of 1000 people increase by 5% in an hour near to accident place. How many people will be in the crowd after 3 hour?

Given:

c=1000

p =1+r =1+0.05=1.05

t =3 hour

Solution:

Exponential function growth g=c(p)^t

Substitute the given data in the formulae

g=1000(1.05)^3

g=1000(1.1576)

g=1157.6

g=1158 people

Exponential Function Decay:

Exponential function decay `d=c(p)^t`

where,

c-Number at initial

p-growth factor

(p=1-r,here r is given decay percentage)

t-time

Example 2: Relevance and application of exponential functions in real life situations

The price of a violin is  $1,000 which decreases at a rate of interest of 3%.What is the price of a violin after 2 year?

Given:

c=1,000

p=1-r=1-0.03=0.97

t=2 year

Solution:

Exponential function decay `d=c(p)^t`

`d=1,000(0.97)^2`

d=1,000(0.9409)

d=940.9

d=941

The price of a violin after 2 year=$941

Example 3: Relevance and application of exponential functions in real life situations

John invests $50,000 for 2 year with the interest of 4% compounded half yearly .Find out the compound interest for his investment at a given rate of interest?

Given:

P=$50,000

R=4%

N=2 year

Solution:

Compound interest =` P[1+(R/2)/100]^(2n)`

`=50,000[1+(4/2)/(100)]4`

`=50,000[1+(2/100)]4`

`=50,000xx(102/100)xx(102/100) xx(102/100) xx(102/100)`

=$50,000(1.0824)

=$54121.6

Compound interest=Total amount-principle

=54,121.6-50,000

=$4,121.6

Example 4: Relevance and application of exponential functions in real life situations

Calculate how long blood clotting cell will take to produce 20,400

Solution:

We need to consider  assume function to calculate population based on problems

`f(t)=2^t`

`20,400=2^t`

Find the natural logarithm on both side

`ln(20,400)=ln(2^t)`

ln(20,400)= t ln(2)

`t= ln(20,400)/ ln(2)`

`t=(9.923)/0.693=14.29`

After 14.32 min the blood clotting cell can produce 20,400  cell.