A mechanical engineer earned a yearly salary of $50,000 in 1990 and has averaged a 6.2% raise anually for the last 10 years and expects that this increase will continue.

a) Write an equation that models this situation. Let S=yearly salary and n=number of yrs. since 1990.

b) According to your equation, what was the engineer's salary in 1980?

c) How long will it take for the engineer's salary to reach $100,000?

d) Write a doubling time equation for this situation.

AdamTaurus Dec 7, 2017

#1**+1 **

A mechanical engineer earned a yearly salary of $50,000 in 1990 and has averaged a 6.2% raise anually for the last 10 years and expects that this increase will continue.

**a)** Write an equation that models this situation.

Let S = yearly salary and n = number of years since 1990 .

when n = 0 , S = 50000

when n = 1 , S = 1.062 * 50000

when n = 2 , S = 1.062 * 1.062 * 50000

when n = 3 , S = 1.062^{3} * 50000

So...

S = 1.062^{n} * 50000

**b)** 1980 = 1990 + -10 , so our n = -10 .

S = 1.062^{-10} * 50000

S ≈ 27398

**c)**

100000 = 1.062^{n} * 50000 Divide both sides by 50000 .

2 = 1.062^{n} Take the ln of both sides.

ln 2 = ln 1.062^{n}

ln 2 = n ln 1.062

ln 2 / ln 1.062 = n

11.5 ≈ n

If it is a steady increase, it will take about 11.5 years after 1990 for the salary to reach 100,000 .

**d)** I don't know what that is...sorry!!!

hectictar Dec 8, 2017

#2**+2 **

d. The doubling time, t, is independent of any amount

So

2A = A (1.062)^t divide both sides by A

2 = 1.062^t take the log of both sides

log 2 = log (1.062)^t and we can write

log (2) = t * log (1.062) divide both sides by log (1.062)

log (2) / log (1.062) = t ≈ 11.52 yrs ≈ 12 years .i.e., it takes any starting amount about 12 years to double

CPhill Dec 8, 2017