## College Algebra (6th Edition)

a) The population in California in 2010 was $37.3$ million. b) It will take around $8$ years for the population to reach $40$ million in the year $2018$.
a) $A = 37.3e^{0.0095t}$ $A = 37.3e^{0.0095(0)}$ $A = 37.3e^{0}$ $A = 37.3(1)$ $A = 37.3$ million The population in California in 2010 was $37.3$ million. b) $40 = 37.3e^{0.0095t}$ $\frac{40}{37.3} = e^{0.0095t}$ $\ln (\frac{40}{37.3}) = 0.0095t$ $t = \frac{(\ln \frac{40}{37.3})}{0.0095}$ $t = 7.3564...$ years (If its $7$ years then the population will be less than $40$ million, but $8$ years will be over $40$ million.) $t \approx 8$years It will take around $8$ years for the population to reach $40$ million in the year $2018$.