Question 393385
1.A biologist is researching a newly-discovered species of bacteria.
 at time t=0 hours, he puts one hundred bacteria into what he has determined to
 be a favorable growth medium.
 Six hours later, he measures 450 bacteria.
 Find an exponential equation that approximates the information.
:
find k, the growth factor
100*(2^(6/k)) = 450
Divide both sides by 100
2^(6/k) = 4.5
use logs here
log(2^(6/k)) = log(4.5)
:
{{{6/k}}}log(2) = log(4.5)
{{{6/k}}} = {{{log(4.5)/log(2)}}}
{{{6/k}}} = 2.17
k = {{{6/2.17}}}
k = 2.765 is the growth factor
:
A = Ao*2^(t/2.765), the exponential equation
where
Ao = initial amt
A = amt after t hrs
t = time in hrs
:
Check equation using Ao = 100: In a calc enter: 100*2^(6/2.765) results: 450.0
:
:
2. A $1,000 deposit is made at a bank that pays 12% compound monthly.
 How much will you have in your account at the end of 10 years.
:
Compound interest formula:
A ={{{p(1+r/n)^(nt))}}}
where 
A = resulting amt after t years
P = principal ($1000)
r = interest rate in decimals (.12)
n = no. of time compounded per yr (12 times/yr)
t = no. of yrs (10 yrs)
:
A ={{{1000(1+.12/12)^(12*10))}}}
:
A ={{{1000(1+.01)^(120))}}}
:
A ={{{1000(1.01)^(120))}}}
on a calc
A = $3300.39