+109 The number of hours, t, that bacteria spread 10-fold can be modeled by the equation B(t) = B0(10)^5t. There are 25 bacteria present initially and a biologist wishes to find out how many hours will elapse until there are 5,000 bacteria present. What is the exact value for the number of hours elapsed, t, in the equation 5,000 = 25(10)^5t?
A.) t= log20 / 5
B.) t= log5,000 / 5log250
C.) t= log200 / log5
D.) t= log 200 / 5
Solve for t over the real numbers:
5000 = 2^(5 t) 5^(5 t+2)
5000 = 2^(5 t) 5^(5 t+2) is equivalent to 2^(5 t) 5^(5 t+2) = 5000:
2^(5 t) 5^(5 t+2) = 5000
Take the natural logarithm of both sides and use the identities log(a b) = log(a)+log(b) and log(a^b) = b log(a):
5 log(2) t+log(5) (5 t+2) = log(5000)
Expand and collect in terms of t:
(5 log(2)+5 log(5)) t+2 log(5) = log(5000)
Subtract 2 log(5) from both sides:
(5 log(2)+5 log(5)) t = log(5000)-2 log(5)
Divide both sides by 5 log(2)+5 log(5):
Answer: | t = (log(5000)-2 log(5))/(5 log(2)+5 log(5)) =Log(200)/5
