CALCULUS exponential function help plz ;/?

Question

the population of a certain kind of bacteria in one cubic centimeter in the blood of a sick person was 1000 in the first day of illness(t=0). if the bacteria is increasing exponentially at a rate of 25% , find :

a) a formula for the number of bacteria in one cubic centimeter,q , as a function of the number of the days t, since the first day of illness.

b)the number of bacteria in one cubic centimeter in the third day of illness.

c)in which day of illness the number of bacteria in one cubic centimeter exceeds 3000 .

Answer 1

a. q(t) = 1000e^.25t

b. q(2) = 1000e^(.25)(2) = 1648

c. 3000 = 1000e^.25t
3 = e^.25t
ln 3 = .25t

ln3/.25 = t

4.39 = t, so the fifth day.

Answer 2

The no of bacteria will be given by the formula q(t) = q0*e^(t/tc); You need to find tc. At t=0, q=q0, which was 1000, so q0 = 1000. I assume the 25% means 25% per day. So for t = 1 day, q=1.25*q0.
Then 1.25*q0 = q0*e^1/tc), or 1.25 = e^1/tc; take ln of both sides to get ln(1.25) = 1/tc. and tc = 1/ln(1.25). Thus the final formula is

q(t) = 1000e^[t/ln(1.25)]

You plug in any value of t (days) you need q for.

For the last part, use 3000 = 1000e^[t/ln(1.25)] and solve for t by taking ln of both sides again.

Answer 3

the spinoff of e^x is e^x so u use the quotient rule, it extremely is [(u' v)-(v' u)]/ v^2 u=e^x v=2+4x g'(x) = [(e^x)(2+4x) - (4)(e^x)]/ (2+4x)^2 = [(e^x)(2+4x-4)]/(2+4x)^2 i in my opinion component out the e^x = [(e^x)(4x-2)]/(2+4x)^2