**Exponential growth and decay word problems :**

To solve exponential growth and decay word problems, we have to be aware of exponential growth and decay functions.

Let us consider the following two examples.

When we invest some money in a bank, it grows year by year, because of the interest paid by the bank.

We buy a car and use it for some years. When it becomes too old, we would like to sell it.

In the first example, we will be keen to know the final value (Amount invested + Interest) of our deposit. To know the final value of the deposit, we have to use growth function.

In the second example, we will be eager to know the sale value of the car (Purchased price - depreciation). Here we have to use decay function.

In this way, growth and decay functions are being used in our life.

Many real world phenomena are being modeled by functions which describe how things grow or decay as time passes.

Let us see the functions which use to estimate and growth and decay.

**Formula 1 :**

The formula given below is related to compound interest formula and represents the case where interest is being compounded continuously.

That is, at any instant the balance is changing at a rate that equals "r" times the current balance.

We use this formula, when it is given "exponential growth/or decay"

A ---> Ending amount

P ---> Beginning amount

r ---> Growth/Decay rate

t ---> Time

**Note :**

If it is decay function, the value of "r" will be negative.

**Formula 2 :**

The formula given below is compound interest formula and represents the case where interest is being compounded annually or the growth is being compounded once the term is completed.

A ---> Ending amount

P ---> Beginning amount

r ---> Growth/Decay rate

n ---> No. of years / Time

**Note :**

If it is decay function, the value of "r" will be negative.

**Formula 3 :**

The formula given below is related to geometric progression. Here, the initial amount will grow/decay at the constant ratio "b".

A ---> Ending amount

a ---> Beginning amount

b ---> Growth/Decay ratio

x ---> No. of years / terms

**Note :**

If it is growth function, we will have "r" > 1

If it is decay function, we will have 0 < r < 1

To have better understanding on "Exponential growth and decay word problems", let us look at some examples.

**Example 1 :**

David owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is 8% annually, how many stores does the restaurant operate in 2007 ?

**Solution :**

Number of years between 1999 and 2007 is

n = 2007 - 1999 = 8

No. of stores in the year 2007 = P(1+r)ⁿ

Here, P = 200, r = 8% or 0.08, n = 8

No. of stores in the year 2007 = 200(1+0.08)⁸

No. of stores in the year 2007 = 200(1.08)⁸

No. of stores in the year 2007 = 200(1+0.08)⁸

No. of stores in the year 2007 = 200(1.8509)

No. of stores in the year 2007 = 370.18

Hence, the number of stores in the year 2007 is 370 (approximately)

**Example 2 :**

You invest $2500 in bank which pays 10% interest per year compounded continuously. What will be the value of the investment after 10 years ?

**Solution :**

We have to use the formula given below to know the value of the investment after 3 years.

Here,

A = Final value of the deposit

P = 2500, r = 10% or 0.1, t = 10, e = 2.71828 and also

rt = 0.1x10 = 1

A = 2500(2.71828)¹

A = 6795.70

Hence, the value of the investment after 10 years is $6795.70

**Example 3 :**

Suppose a radio active substance decays at a rate of 3.5% per hour. What percent of substance will be left after 6 hours ?

**Solution :**

Since the initial amount of substance is not given and the problem is based on percentage, we have to assume that the initial amount of substance is 100.

We have to use the formula given below to find the percent of substance after 6 hours.

Here,

A = Amount of substance after 6 hours.

P = 100, r = -3.5% or -0.035, t = 6

(Here, the value of "r" is taken in negative sign. because the substance decays)

A = 100(1-0.035)⁶

A = 100(0.935)⁶

A = 100(0.8075)

A = 80.75

Since the initial amount of substance is assumed as 100, the percent of substance left after 6 hours is 80.75%

**Example 4 :**

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture initially, how many bacteria will be present at the end of 8th hour?

**Solution :**

Note that the number of bacteria present in the culture doubles at the end of successive hours.

Since it grows at the constant ratio "2", the growth is based is on geometric progression.

We have to use the formula given below to find the no. of bacteria present at the end of 8th hour.

Here,

A = No. of bacteria at the end of 8th hour

a = 30, b = 2 and x = 8

A = 30x2⁸

A = 30x256

A = 7680

Hence, the number of bacteria at the end of 8th hour is 7680.

**Example 5 :**

A sum of money placed at compound interest doubles itself in 3 years. If interest is being compounded annually, in how many years will it amount to four times itself ?

**Solution :**

Let "P" be the amount invested initially.

From the given information, P becomes 2P in 3 years.

Since the investment is in compound interest, for the 4th year, the principal will be 2P.

And 2P becomes 4P (it doubles itself) in the next 3 years.

Therefore, at the end of 6 years accumulated value will be 4P.

Hence, the amount deposited will amount to 4 times itself in 6 years.

After having gone through the stuff given above, we hope that the students would have understood "Exponential growth and decay word problems".

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