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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
avatar+9460 
+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  =  1.0623 * 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
 #2
avatar+128079 
+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

 

 

cool cool cool

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

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