Question 1198712
a) If the city grows at 3.87% per year, that is equivalent to multiplying by 1.0387. Therefore, the equation is {{{P=6970000*1.0387^t}}}, where {{{P}}} is the population, and {{{t}}} is the number of years after 2012. 
b) 2018 is 6 years after 2012, so we would have {{{t=6}}}. Plugging this into our equation, we get {{{P=6970000*1.0387^6}}}. We can simply plug this into a calculator to get {{{P=highlight(8753335)}}}.
c) Since {{{P}}} is the population, we can plug in 10 million for {{{P}}} and solve for {{{t}}}.
We have the equation {{{10000000=6970000*1.0387^t}}}. Dividing both sides by {{{6970000}}}, we get {{{1.0387^t=1.43472023}}}. Taking the natural log of both sides, we get {{{ln(1.0387^t)=ln(1.43472023)}}}. Using logarithm rules to take the exponent out, we get {{{t*ln(1.0387)=ln(1.43472023)}}}. Dividing both sides by {{{ln(1.0387)}}}, we get {{{t=ln(1.43472023)/ln(1.0387)}}}. Plugging this into a calculator, we get {{{t=9.50672957}}}. Since {{{t}}} is the number of years after 2012, the population will be 10 million in 2021.
d) To find the doubling time, we can solve the equation {{{1.0387^t=2}}}, since {{{1.0387^t}}} is how much the population is growing. We can take the natural log of both sides to get {{{ln(1.0387^t)=ln(2)}}}. We can use logarithm rules to take out the exponent, which will give {{{t*ln(1.0387)=ln(2)}}}. We can divide both sides by {{{ln(1.0387)}}} to get {{{t=ln(2)/ln(1.0387)}}}. Finally, we can plug this into a calculator to get {{{t=highlight(18.25516027)}}}.