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Assume that $5000 can be invested at rate of 5%/a compounded monthly

a.) determine the value of the investment after 6 years

A=Ao(1+i)n

A=5000(1.05)6

=6700$

b.) determione the length of time to the nearest year, required for the investement to tripple in value ?

 Jun 5, 2014

Best Answer 

 #1
avatar+129850 
+10

Your set-up to "a" is slightly incorrect...it should be

A = 5000(1+.05/12)12*6  =  $6745.09

 

For "b', the amount invested doesn't really matter....let's see why......

3A = A (1 + .05/12)12t      divide through by A....

3 = (1 + .05/12)12t       (see why "A" doesn't matter??) .......take the log of both sides

log3 = log (1 + .05/12)12t         and by a property of logs, we can write

log3 = (12t)*log(1 + .05/12)       divide both sides by 12*log(1 + .05/12)

[log3] / (12*log(1 + .05/12)) = t = about 22.018 years

 

 Jun 5, 2014
 #1
avatar+129850 
+10
Best Answer

Your set-up to "a" is slightly incorrect...it should be

A = 5000(1+.05/12)12*6  =  $6745.09

 

For "b', the amount invested doesn't really matter....let's see why......

3A = A (1 + .05/12)12t      divide through by A....

3 = (1 + .05/12)12t       (see why "A" doesn't matter??) .......take the log of both sides

log3 = log (1 + .05/12)12t         and by a property of logs, we can write

log3 = (12t)*log(1 + .05/12)       divide both sides by 12*log(1 + .05/12)

[log3] / (12*log(1 + .05/12)) = t = about 22.018 years

 

CPhill Jun 5, 2014

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