SOLUTION: Bob has $8300.00 he invests at a rate of 6.2% interest per year compounded quarterly. a) How much does Bob have after 3 and 1/2 years ? b) How long will it take for Bob's investm

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Bob has $8300.00 he invests at a rate of 6.2% interest per year compounded quarterly. a) How much does Bob have after 3 and 1/2 years ? b) How long will it take for Bob's investm      Log On


   

Question 1096088: Bob has $8300.00 he invests at a rate of 6.2% interest per year compounded quarterly.
a) How much does Bob have after 3 and 1/2 years ?
b) How long will it take for Bob's investment to reach $15,000.00 ?
Answer by VFBundy(438) About Me  (Show Source):
You can put this solution on YOUR website!
This is the formula for compound interest:

N=P%281%2Br%2Fn%29%5Ent

N is the new amount.
P is the principal amount.
r is the annual interest rate.
n is how often the interest is compounded per year.
t is the number of years.

a) How much does Bob have after 3 and 1/2 years ?
N=8300%281%2B0.062%2F4%29%5E%284%2A3.5%29 = $10,294.31

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b) How long will it take for Bob's investment to reach $15,000.00 ?
15000=8300%281%2B0.062%2F4%29%5E%284t%29
Divide each side of the equation by 8300:
1.8072=%281%2B0.062%2F4%29%5E%284t%29
Simplify the stuff inside the parenthesis:
1.8072=%281.0155%29%5E%284t%29

Convert this to log form:
log 1.8072 base 1.0155 = 4t

Compute log 1.8072 base 1.0155:
38.4744 = 4t

Solve for t:
t = 9.6186

It will take 9.6186 years for Bob's investment to reach $15,000.