Question

The population of a certain country in 1996 was 286 million people. In​ addition, the population of the country was growing at a rate of 0.8​% per year. Assuming that this growth rate​ continues, the model P(t) = 286(1.008 )^t-1996 represents the population P​ (in millions of​ people) in year t. According to this​ model, when will the population of the country reach A. 306 million? B. 386 million?

Answer 1

Answer:

A) In 2004 the population will reach 306 million.

B) In 2033 the population will reach 386 million.

Step-by-step explanation:

Given : The population of a certain country in 1996 was 286 million people. In​ addition, the population of the country was growing at a rate of 0.8​% per year. Assuming that this growth rate​ continues, the model $P(t) = 286(1.008 )^{t-1996}$ represents the population P​ (in millions of​ people) in year t.

To find : According to this​ model, when will the population of the country reach A. 306 million? B. 386 million?

Solution :

The model represent the population is $P(t) = 286(1.008 )^{t-1996}$

Where, P represents the population in million.

t represents the time.

A) When population P=306 million.

$306 = 286(1.008 )^{t-1996}$

$\frac{306}{286}=(1.008 )^{t-1996}$

$1.0699=(1.008 )^{t-1996}$

Taking log both side,

$\log(1.0699)=\log((1.008 )^{t-1996})$

$\log(1.0699)=(t-1996)\log(1.008)$  

$\frac{\log(1.0699)}{\log(1.008)}=(t-1996)$  

$8.479=t-1996$  

$t=8.479+1996$

$t=2004.47$

$t\approx2004$

Therefore, In 2004 the population will reach 306 million.

B) When population P=386 million.

$386 = 286(1.008 )^{t-1996}$

$\frac{386}{286}=(1.008 )^{t-1996}$

$1.3496=(1.008 )^{t-1996}$

Taking log both side,

$\log(1.3496)=\log((1.008 )^{t-1996})$

$\log(1.3496)=(t-1996)\log(1.008)$  

$\frac{\log(1.3496)}{\log(1.008)}=(t-1996)$  

$37.625=t-1996$  

$t=37.625+1996$

$t=2033.625$

$t\approx2033$

Therefore, In 2033 the population will reach 386 million.