A bacteria culture grows with constant relative growth rate. The bacteria count was 1296 after 2 hours and 531441 after 6 hours. Answer the following questions. 1. What is the relative growth rate

A bacteria culture grows with constant relative growth rate. The bacteria count was 1296 after 2 hours and 531441 after 6 hours. Answer the following questions.

1. What is the relative growth rate?

2. What was the initial size of the culture?

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

4. Find the number of cells after 5 hours.

5. When will the population reach 1000000?

To find relative growth rate, expression below can be used

P(t)= Po*e^(kt)

Where k= growth rate

Po= initial amount

t= time in hours

We were given The bacteria count as 1296 after 2 hours and 531441 after 6 hours. This can be expressed below as

P(2)= P(o)e^(2k)= 1296.........eqn(1)

P(6)= P(o)e^(6k)= 531441........eqn(2)

If we find the ratio of the two eqn we have

531441/1296 =[P(o)e^(6k)]/[P(o)e^(2k)]

410.0625= e^(4k)

If we take Ln of both side

ln 410.0625=ln(e^4k)

k=1.50408

1)= k=1.50408

2)

From P(2)= Po*e^(2k)= 1296

1296=Po × e^(2×1.50408)

1296= Po × e^(3.00816)

Po=1296/e^(3.00816)

=64

3) the expression is P(t)=64e^(1.50408t)

4). P(5)=64e^(1.50408×5)

=118098

5) P(t)= 1,000,000

1000000=64e^(1.50408t)

1000000/64 = 64e^(1.50408t)

Ln(15625)=ln[e^(1.50408t)]

9.657=e^(1.50408t)

t=6.42 hrs