Question 251300
assume his salary starts at 1.

at the beginning of the first year he is making 1.
at the end of the first year he is making 1 * 1.1 = 1.1
at the end of the second year he is making 1.1 * 1.2 = 1.32
at the end of the third year he is making 1.32 * 1.25 = 1.65


he starts at 1 and at the end of the third year he is making 1.65


his average annual increase is given by the compounding formula of:


1.65 = 1*(1+x)^3


this is equivalent to:


1.65 = (1+x)^3


take the cube root of both sides to get:


(1.65)^(1/3) = 1+x


subtract 1 from both sides to get:


x = (1.65)^(1/3) - 1


solve for x to get:


x = .18166575


at the beginning of the first year he is making 1.
at the end of the first year he is making 1 * 1.18166575 = 1.18166575
at the end of the second year he is making 1.18166575 * 1.18166575 = 1.396333946
at the endof the third year he is making 1.396333946 * 1.18166575 = 165


1 * (1.18166575)^3 = 1.65


the equation that was used was the future value of a present amount formula that is equal to:


f = p * (1+i)^n


f = future value
p = present amount
i = interest rate per time period
n = number of time periods.


to solve this problem, we first had to find f.


that was done using the year by year analysis up top.


once we knew f, we could then substitute in the formula to get:


f = 1.65
p = 1
i = x
n = 3


we then solved for x.