Answer:
a) growth
b) 3%
c) 6 years (since the beginning of the decade)
Explanation:
Given:
The population P (in thousands) of Austin, Texas during a recent decade can be approximated by
when t is the number of years since the beginning of the decade.
General form of an exponential function:
![y=ab^x](https://img.qammunity.org/2023/formulas/mathematics/high-school/hye5rg1h8wj3ohgdt4j1vpepdhoym0w9ex.png)
where:
- a is the y-intercept (or initial value)
- b is the base (or growth factor)
- x is the independent variable
- y is the dependent variable
If
then it is an increasing function
If
then it is a decreasing function
a) The model represents exponential growth as 1.03 > 1
b) The annual percent increase of the population is 3%
1.03 - 1 = 0.03
0.03 x 100 = 3%
c) To estimate when was population about 590,000 set y = 590 and solve for t:
![\implies 590=494.29(1.03)^t](https://img.qammunity.org/2023/formulas/mathematics/high-school/nmptgbmd9yc5ykb32rv4agth9xyr3m8aky.png)
![\implies (590)/(494.29)=(1.03)^t](https://img.qammunity.org/2023/formulas/mathematics/high-school/iwnehpmmu33evl2jnoxrgkdzu5wua5zlb8.png)
Take natural logs of both sides:
![\implies \ln\left((590)/(494.29)\right)=\ln (1.03)^t](https://img.qammunity.org/2023/formulas/mathematics/high-school/8kl5f084ubftyze2i9ljpwakmgvula41te.png)
![\implies \ln\left((590)/(494.29)\right)=t\ln (1.03)](https://img.qammunity.org/2023/formulas/mathematics/high-school/lccl74l8k1wovjf221y1im1qg9bxx6293a.png)
![\implies t=(\ln\left((590)/(494.29)\right))/(\ln (1.03))](https://img.qammunity.org/2023/formulas/mathematics/high-school/qy2w4a7130zuz6hpo4dtbatr8cyof5ayq4.png)
![\implies t=5.988069001...](https://img.qammunity.org/2023/formulas/mathematics/high-school/m5j3icwan2gbeg1f8i9y6hlff6gxel2ceb.png)
Therefore the population was about 590,000 6 years since the beginning of the decade.