# Exponential growth

• December 9th 2009, 09:15 AM
DBA
Exponential growth
A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420.

a) Find an expression for the number of bacteria after t hours.

The formula I used is

P(t) = P(0) * e^(k*t)

Step 1
I need P(0) --> We know that at t=0 P=100
So P(0) = 100

Step 2
I need k

I have
420 = 100 * e^(k*1)
100 = 100 * e^(k*0)

I took the ratio

420/100 = 100 * e^(k*1) / 100 = 100 * e^(k*0)

4.2 = e^k
ln 4.2 = ln e^k
ln 4.2 = k* lne
ln 4.2 = k* 1

So k=ln4.2

Step 3
Write the expression by plugging in P(0) and k

P(t) = 100 * e^[(ln4.2) *t]

The answer in the book is P(t) = 100 * (ln4.2)^t

I do not understand why

e^[(ln4.2) *t] = (ln4.2)^t

Can someone explain that to me please?
Thanks
• December 9th 2009, 09:21 AM
e^(i*pi)
Quote:

Originally Posted by DBA
A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420.

a) Find an expression for the number of bacteria after t hours.

The formula I used is

P(t) = P(0) * e^(k*t)

Step 1
I need P(0) --> We know that at t=0 P=100
So P(0) = 100

e^(i*pi) - correct

Step 2
I need k

I have
420 = 100 * e^(k*1)
100 = 100 * e^(k*0)

I took the ratio

420/100 = 100 * e^(k*1) / 100 = 100 * e^(k*0)

4.2 = e^k
ln 4.2 = ln e^k
ln 4.2 = k* lne
ln 4.2 = k* 1

So k=ln4.2

e^(i*pi) - correct (although the second equation is superfluous)

Step 3
Write the expression by plugging in P(0) and k

P(t) = 100 * e^[(ln4.2) *t]

e^(i*pi) - correct

The answer in the book is P(t) = 100 * (ln4.2)^t

I do not understand why

e^[(ln4.2) *t] = (ln4.2)^t

Can someone explain that to me please?
Thanks

Those two expression are not equal, as your working seems fine I would say the book is in error

You can simplify using the log power law

$e^{ln(4.2) \cdot t} = e^{ln(4.2^t)} = 4.2^t$

$\therefore \: \: P(t) = 100 \cdot 4.2^t$