Problema Solution
The population of a city in 1975 was 30,000. In 1978, the population was 39,000. Assuming that y=y0e^kt (zero is the subscript of y) find the value of k.
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y(t) = y0e^(kt), where t is the time in years after 1975
we will assume that y0 is the amount of the substance present at time t=0, and that y0 = 30,000
y(0) = y0e^(k*0)
y0e^(k*0) = 30,000
y0 = 30,000
in 1978 the value for t is:
t = 1978 - 1975 = 3
y(3) = 39,000 that is
y0e^(k*3) = 39,000
30,000 e^(3k) = 39,000 divide both sides by 30,000
e^(3k) = 39,000/30,000
e^(3k) = 39/30
3k = ln(13/20)
k = (1/3)(ln(13/20))
k = -0.1436
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