Q:

# The exponential model Upper A equals 104.8 e Superscript 0.001 t describes the​ population, A, of a country in​ millions, t years after 2003. Use the model to determine when the population of the country will be 106 million.

Accepted Solution

A:
Answer:  The population of the country will be 106 millions in 2014. Step-by-step explanation: The exercise gives you the following exponential model, which describes the​ population "A" (in​ millions) of a country "t" years after 2003: $$A=104.8 e^{0.001 t}$$ In this case you must determine when the population of that country will be 106 millions, so you can identify that: $$A=106$$ Now you need to substitute this value into the exponential model given in the exercise: $$106=104.8 e^{0.001 t}$$ Finally, you must solve for "t", but first it is important to remember the following Properties of logarithms: $$ln(a)^b=b*ln(a)\\\\ln(e)=1$$ Then: $$\frac{106}{104.8}=e^{0.001 t}\\\\ln(\frac{106}{104.8})=ln(e)^{0.001 t}\\\\ln(\frac{106}{104.8})=0.001 t(1)\\\\\frac{ln(\frac{106}{104.8})}{0.001}}=t\\\\t=11.38\\\\t\approx11$$ Notice that in 11 years the population will be 106 millions, then the year will be: $$2003+11=2014$$ The population of the country will be 106 millions in 2014.