Exponential Growth Formula with Solved Examples

Ans. A = Pert

Substitute 

P = 2500

r = 10% or 0.1

t = 10

e = 2.71828

Then, we have, A = 2500(2.71828)^(0.1)10

A = 6795.70

So, the value of the investment after 10 years is Rs 6795.70. 

Ans. If X (t) = X0 e^rt 

Given is X_o = 5

r= 4%, which is 0.04

t = 15 years

Therefore X (15) = 5 x e0.04 x 15

Substituting Euler’s number, X (15) = 9.11059 million

Therefore the population in 15 years will be 9.11059 million. 

Ans. 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 on geometric progression. 

Use the formula given below to find the number of bacteria present at the end of the 8th hour. 

A = ab^x

Substitute, a = 30

b = 2

x = 8

Then, we have

A = 30(2⁸)

A = 30(256)

A = 7680

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

Ans. Here there is no direct mention of differential equations, but the use of ‘growing exponentially’ must be taken as an indicator that we are talking about the situation

f(t)=ce^kt

If f(t) is the number of llamas at time t and c,k are constants to be determined from the information given in the problem. And the use of language should probably be taken to mean that at time t=0 there are 1000 llamas, and at time t=4 there are 2000.

Then, either repeating the method above or plugging into the formula derived by the method, we find

c= value of f at t=0 = 1000

k= lnf(t¹)−lnf(t²) / t¹−t²

=ln1000−ln2000 / 0−4

= ln (1000 / 2000) – 4

= ln (1/2) / -4

= ( ln 2)/4

Therefore, f(t) = 1000 e ^{((ln2)/4) t} =1000⋅2^t/4

This is the desired formula for the number of llamas at arbitrary time t. 

Ans. We need to find the general formula for the number f(t) of bacteria at time t, and  set this expression equal to 100,000 to solve t. We can take a little shortcut here since we know that c=f(0) and we are given that f(0)=10f. (This is easier than using the bulkier more general formula for finding c). And use the formula for k:

K = lnf(t¹)−lnf(t²) / t¹−t²

= ln10−ln²,000 / 0−4

= (ln(10/2,000))/ -4

= ln 200/ 4

Therefore, we have   f(t)=10⋅e ^{(ln200/4) t} = 10⋅200t/4

 as the general formula. Now we try to solve

100,000=10⋅e(ln200/4) t

for t: divide both sides by the 10 and take logarithms, to get

ln10,000= (ln200/4) t

Thus, t=4(ln10,000/ln200) ≈6.953407835.

Ans. Rewrite the above as

y = x – π

But then dy/dx = 1

Alternatively, directly differentiating the relationship w.r.t., x, we have

d/ dx (x – y)  = d π/ dx

d π/ dx means to differentiate the constant function taking value π everywhere w.r.t., x. Thus

d/dx(x) − d/ dx (y) = 0

which is equal to , dy/dx = dx/dx =1.

Ans. Start with the formula: y(t) = a × e^kt

We know a=3 mice, t=2 months, and right now y(2)=18 mice: 18 = 3 × e^2k    

Now some algebra to solve for k, divide both sides by 3:6 = e^2k

Take the natural Logarithm of both sides: ln(6) = ln(e^2k)

ln(ex)=x, so:ln(6) = 2k

2k = ln(6)

Divide by 2:k = ln(6)/2

We can now put k = ln(6)/2 into our formula from before:

y(t) = 3 e^(ln(6)/2)t

Now let's calculate the population in 2 more months (at t=4 months):

y(4) = 3 e^(ln(6)/2)×4 = 108

And in 1 year from now (t=14 months):

y(14) = 3 e^{(ln(6)/2)×14} = 839,808 

Ans. y(t) = a × ekt

We know:

• a (the starting dose) = 1 cup of coffee!
• t is in hours
• at y(6) we have a 50% reduction  (6 is the half-life)

So, 0.5 = 1 cup × e^6k

Now some algebra to solve for k:

Take the natural logarithm of both sides: ln(0.5) = ln(e^6k)

 ln(ex)=x, so:ln(0.5) = 6k

 6k = ln(0.5)

 Divide by 6:k = ln(0.5)/6

Now we can write: y(t) = 1 e^{(ln(0.5)/6)×t}

In 6 hours: y(6) = 1e^{(ln(0.5)/6)×6} = 0.5

And in 9 hours: y(9) = 1e^{(ln(0.5)/6)×9} = 0.35

After 9 hours the amount left in your system is about 0.35 of the original amount.