Exponential Population Growth Graph

In this article we will learn about exponential population growth graph. An Exponential functions is involve exponents, basically it deals with such function f(x) = x^4 where the variable x was the base and the number 4 was the power. Exponent is how many times to use the number in a multiplication like 23 = 2 *2 * 2.   Let us see Exponential population growth graph.
Exponential Function:

An exponential function has the form f(x) = a.b^x or b>0 , where a is the initial value and b is the base. There are two types of exponential functions:

I like to share this Exponential Calculator with you all through my article.

Exponential growth tutoring:

An exponential function of the form f(x) = a.b^x is an exponential growth function while b>1.
Each time x is increased by 1, y automatically decreases to one half of its previous value.
An Exponential growth is increase in a quantity. Therefore the function is y = a( 1+ r)^x

a = initial amount before measuring growth

r = growth rate

x = number of time intervals.  Let us see Exponential population growth graph.
Exponential Population Growth Graph:

Example 1: Exponential population growth graph

In 2000, there were 200 toys sold in the small town.  The number of sales increased by 85% per year after 2000.  How many toys were sold in the year 2004?

Solution:

Step1:

Given: Here a = 200

r = 85% = .85

x = number of time intervals ( 1 , 2 , 3 , 4 )

Step2: plug the values in  y = a( 1+ r)^x

y = 200(1+.85)^1    = 200 (1.85)^1   = 200*1.85 = 370

y = 200(1+.85)^2 = 200 (1.85)^2   = 200*3.4225 = 684.5

y = 200(1+.85)^3    = 200 (1.85)^3   = 200*6.331625 = 1266.325

y = 200(1+.85)^4   = 200 (1.85)^4   = 200*11.71350625 = 2342.70125

Step 3: Now plots the points on graphs

Exponential population growth graph

Example 2: Exponential population growth graph

The population of a city is p = 450,300 e^(0.6t) here t = 0 is denoted the population in the year 2000.find the population of the city in year 2005?

Solution:

Now we should find population in the year 2005,

So, let t = 5 in given equation

P = 450300 e^(0.6(5))  = 450300 e^3 = 450300 * 20.0855 =9044500.65

Therefore in 2005 the population of city near 9044500 people.

I am planning to write more post on Definition of Adjacent Angles and factoring polynomials word problems. Keep checking my blog.

Example 3: Exponential population growth graph

In 2000, there were 200256 people in the small town. Population increased 50% per after the year 2000. Find how many people in the year 2004?

Solution:

Step1:

Given: Here a = 200256

r = 50% = 0.5

x = number of time intervals ( 1 , 2 , 3 , 4 )

Step2: plug the values in  y = a( 1+ r)^x

y = 200256(1+.5)^1    = 200256 (1.5)^1   = 200256*1.5 = 300384

y = 200(1+.5)^2   = 200256 (1.5)^2   = 200256*2.25 = 450576

y = 200(1+.5)^3    = 200256 (1.5)^3   = 200256*3.375 =  675 864

y = 200(1+.5)^4   = 200256 (1.5)^4   = 200256*5.0625= 1013796

Step 3: Now plots the points on graphs

Exponential population growth graph