algebra please help

Question

Suppose a city with population 700,000 has been growing at a rate of 7​% per year. If this rate​ continues, find the population of this city in 11 years.

Answer 1

A=700e^.07(11) = 700e^.77 = 1511.836378 thousand = 1,511,836

population realisticaly grows continuously, rather than compounded annually

if you claimed somewhere somehow that people only increased in number once a year

maybe due to some strictly legally enforced immigration laws

then the formula would be A=700(1+.07)^11 = 1,473,396

the two formulas give roughtly the same numbers, for shorter time periods.

Answer 2

Let's use 000's in the calculation so there will be fewer 0's to contend with.

The problem is exponential growth (just like yearly compound interest in a bank) For this problem, the form is the same, but we can use more specific variables.

P₁₁ = P₀(1 + r)^t where P₀ is the initial population, P₁₁ is the population after 11 years of growth, r is the growth rate as a decimal and t is the number of growth periods

P₁₁ = 700(1 + 0.07)¹¹ → 700(1.07)¹¹ → 700(2.10485) → 1473.3964

So the answer is 1,473,396. Since the input is rounded to the nearest hundred thousand, you may want to round this result as well, as the answer can't be more accurate than the initial value.

You can do this on a regular calculator if you lock in the growth factor of 1.07 (enter 1.07 and hit the "X" key twice), enter 700, then hit the "=" key 11 times.