how do you find the population (in thousands) of people of a city is growing according to the function P(t)=1500(2)^0.144t, where t is the number of years since 2001. I need to find the population of the city in 2001 and in 2008.

2001: t = 0.

P(t) = 1500*(2^)^0 = 1500*(1) = 1500
Thousands.

Year 2008: t = 2008-2001 = 7 yrs.

To find the population of the city in 2001 and 2008 using the given function, you need to substitute the appropriate values of t into the function P(t) and solve for the population.

To find the population in 2001, we substitute t = 0 since it is the number of years since 2001.

P(0) = 1500(2)^0.144(0)
= 1500(2)^0
= 1500(1)
= 1500

So, the population of the city in 2001 is 1500.

To find the population in 2008, we need to find the value of t for the year 2008 since it is the number of years since 2001.

Since 2008 is 7 years after 2001, we substitute t = 7 into the function.

P(7) = 1500(2)^0.144(7)
= 1500(2)^0.1008
≈ 1500(2)^0.101
≈ 1500(1.107)
≈ 1660.5

So, the population of the city in 2008 is approximately 1660.5 thousand.

To find the population of the city in 2001, we will substitute t = 0 into the function P(t) = 1500(2)^(0.144t).

P(0) = 1500(2)^(0.144 * 0)
= 1500(2)^0
= 1500(1)
= 1500

Therefore, the population of the city in 2001 is 1500.

To find the population of the city in 2008, we will substitute t = 7 into the function P(t) = 1500(2)^(0.144t).

P(7) = 1500(2)^(0.144 * 7)
≈ 1500(2)^(0.1)
≈ 1500(1.07177346)
≈ 1607.66

Therefore, the population of the city in 2008 is approximately 1607.66 thousand.