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what rate of interest compounded annually is required to double an investment in 3 years?

 Apr 9, 2015

Best Answer 

 #1
avatar+129842 
+5

We have

2A  =A(1 + r)^3     divide both sides by A

2 = (1 + r)^3         take the log of both sides

log 2  = log(1+ r)^3    and we can write

log 2 =3log(1 + r)    divide both sides by 3

 (log2) /3  = log ( 1 + r)    in exponential form, we have

10 ^[log2 ]/3 = 1 +  r    subtract 1 from both sides

10 ^[log2 ]/3 - 1 = r =  about 25.9 %

 

  

 Apr 9, 2015
 #1
avatar+129842 
+5
Best Answer

We have

2A  =A(1 + r)^3     divide both sides by A

2 = (1 + r)^3         take the log of both sides

log 2  = log(1+ r)^3    and we can write

log 2 =3log(1 + r)    divide both sides by 3

 (log2) /3  = log ( 1 + r)    in exponential form, we have

10 ^[log2 ]/3 = 1 +  r    subtract 1 from both sides

10 ^[log2 ]/3 - 1 = r =  about 25.9 %

 

  

CPhill Apr 9, 2015
 #2
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0

It can be solved more easily.

According to rule 72, the rate needed to double the investment is 72 / 3 = 24

 Oct 30, 2016

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