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# Exponential Functions

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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.

Dec 7, 2017

#1
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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  =  1.0623 * 50000

So...

S  =  1.062n  *  50000

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

S  =  1.062-10  *  50000

S  ≈  27398

c)

100000   =   1.062n  *  50000          Divide both sides by  50000 .

2   =   1.062n                                 Take the  ln  of both sides.

ln 2   =   ln 1.062n

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!!!

Dec 8, 2017
edited by hectictar  Dec 8, 2017
edited by hectictar  Dec 8, 2017
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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

Dec 8, 2017
edited by CPhill  Dec 8, 2017
edited by CPhill  Dec 8, 2017