Compond Interest

Question

How long will it take for an investment of $36,000 to grow to $39,800 if the investment earns interest at the rate of 6%/year compounded annually?

Answer 1

For this you would use the formula n= [ln(V_n/V_p)] / [ln(1+r),  where n is the number of periods (in this case, years), V_n is the future value (39,800) and V_p is the present or starting value (36,000) and r is the rate at which it grows. (note: this formula works best using a TI 30xs calculator) 

So we plug the given info into the formula 
n= [ln(Vn/Vp)] / [ln(1+r)

  = [ln(39,800/36,000)] / [ln(1+6%)]

  = [ln(39,800/36,000)] / [ln(1+.06)]

  = [ln(39,800/36,000)] / [ln(1.06]
  = [0.100347974] / [0.058268908)]
 Which gives us a grand total of 
1.722152985, or about 1.72 years

Answer 2

39800 = 36000(1.06)^t

 

39800/36000 = 1.06^t

 

ln (39800/36000) = t * ln(1.06)

 

t = 1.72 years