
Exponential growth
could anyone give me a hand with these please
1 At the beginning of 1995, an investor decided to invest £6 000 in a plan that grew at an average rate of 6% per year. Show that, if this percentage rate of increase is maintained for 10 years, the value of the original investment will be about £10 745.
2 A population of bacteria is growing exponentially. Explain briefly what this means, using a sketch graph to illustrate your answer.
(i) At time t hours, the size, P, of the population of bacteria is given by P = 1000 (2.8)^0.02t. State the value of P when t = 0. Find the value of t when the size of the population is twice its initial value.
(ii) A second population of bacteria doubles in size every 12 hours. Find the number of hours it takes for this population to increase in size by a factor of 100.

At the beginning of 95,
Principal,
$\displaystyle p = 6000$
Interest,
$\displaystyle I = \frac{6}{100}$
Final Amount at the end of 95,
$\displaystyle A= p + p(I)$
$\displaystyle A= 6000 + 6000(\frac{6}{100})$
$\displaystyle A= 6000(1+\frac{6}{100}) $
At the beginning of 96,
Principal,
$\displaystyle p = 6000(1+\frac{6}{100})$
Interest,
$\displaystyle I = \frac{6}{100}$
Final Amount at the end of 96,
$\displaystyle A= p + p(I)$
$\displaystyle A= 6000(1+\frac{6}{100}) + 6000(1+\frac{6}{100})(\frac{6}{100})$
$\displaystyle A= 6000(1+\frac{6}{100})(1+\frac{6}{100})$
$\displaystyle A= 6000(1+\frac{6}{100})^2$
At the beginning of 97,
Principal,
$\displaystyle p = 6000(1+\frac{6}{100})^2$
Interest,
$\displaystyle I = \frac{6}{100}$
Final Amount at the end of 97,
$\displaystyle A= p + p(I)$
$\displaystyle A= 6000(1+\frac{6}{100})^2 + 6000(1+\frac{6}{100})^2(\frac{6}{100})$
$\displaystyle A= 6000(1+\frac{6}{100})^2(1+\frac{6}{100})$
$\displaystyle A= 6000(1+\frac{6}{100})^3$
So, from here we can safely deduce that after n years, the amount will be
$\displaystyle A= 6000(1+\frac{6}{100})^n$
so after 10 years
$\displaystyle A= 6000(1+\frac{6}{100})^10$
$\displaystyle A= 10745$
This however is the long approach to doing things. If you can identify that it is a geometric progression from the question, you can straight use the $\displaystyle Tn = ar^n$ formula
For the second set of problem
i)
P = 1000(2.8)^0.02t
when t = 0
p = 1000(2.8)^0.02(0)
p = 1000
when p = 2000,
2000 = 1000(2.8)^0.02t
2 = 2.8^0.02t
ln 2 = 0.02t(ln 2.8)
t = ln 2/0.02(ln2.8)
ii)
let the initial population size = X
Population size = (X)(2^n), where n= number of times the bacteria multiplied
Given information, population size is p = 100X
hence,
100X = X(2^n)
n ln 2= ln 100
n = ln 100/ ln 2
n = 6.644 times
therefore time taken is 6.644 x 12 = 79.728 hours