Pre-calculus exponentional Function Aplications continuous interest

M.E.T.S. Pre-Calculus Mr. Mui Name:___________________ Period: ________
Date:________
2.5 “Exponential” Function Applications
Continuously compounded interest rate problems
1) Anthony invests $4,000 at 3% interest compounded continuously. How much money will
he have in 4 Years

http://www.moneychimp.com/articles/finworks/continuous_compounding.htm

A = Pe^(rt) = 4000.00*e^(.03*4) = 4000*e^(.12) = 4509.9874063175

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2) Jose invested $10,000 at 3% interest compounded continuously. How much will he have
after 8 Years?

A = Pe^(rt) = 10000*e^(.03*8) =  10000 * (e ^ (.03 * 8))= 12712.49150321405

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3) Athena wants to double her money. She put $5,000 in a bank account that pays 4%
compounded continuously. How long will it take her to double her money? (Round to the
nearest tenth.)

Time to double continuous compounding video
FORMULA
Amount="pert" Principal e to the Rate Time
          rt
A=Pe
10,000=5,000 e^(.04)t
2=e^.04t 
Take Natural log of both sides
ln 2 = .04t ln (e)  NOTE: ln (e)=1
(ln 2)/.04= t 
0.69314718056/.04=17.328679514 round= 17.3 yr
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4) Michelle invests some money at 2% compounded continuously. If after 6 years she has
$1691.25, what was her initial investment? Calculate Initial Investment Video

2% continuous interest.  after 6 years 1691.25
initial investment=?
A=Pe^(rt)   NOTE: solve for P 

1691.25 = Pe^(.02*6)
P = 1691.25*(e^(-.12))= 1500.00418859789 ROUND to 1500.00 Ans


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5) Shawna invests $2150 at 2% compounded continuously. How many years will it take her
to accumulate $2733.19 in the account?

2150 at 2% continuous How many years to 2733.19 ?
A=Pe^(rt)
2733.19 = 2150e^(.02t)
ln (2733.19) = .02t ln (2150)

ln(2733.19)= 7.91322470418

log(2150) = 7.67322312112

7.91322470418= .02t (7.67322312112)

t =   7.91322470418 / (.02 * 7.67322312112) = 51.56389029272  Round to 51.6 Years

6) Kyara invests $3500 at 2% compounded continuously for 5 years. How much will she
have in her account after 5 years?

A=3500 e^(.02*5)
A=3500e^.1

e ^ .1= 1.10517091808

A = 3500* 1.10517091808 = 3868.09821328 (Round) 3868.09 ans

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7) Zeena invested in an account paying 4% compounded continuously for 3 years. If the
account has $18,039.95 after 3 years, how much did she put in initially?

  18039.95 = Pe^(.04*3)P = 18039.95 (e ^ (-.12))= 16000.00033235568 ROUNDING = 16,000 Ans.
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8) Chelsea put $7500 into an account paying 5% compounded continuously. She now has
$10,643.01. How long has the money been in the account?

  10643.01 = 7500 e^(.05t) 10643.01/7500= e^(.05t)
ln (10643.01/7500)= ln(e^(.05t))=.05t
 ln (10643.01/7500)/.05=t= 7.00000636202 ROUND to 7.0 Years Ans. 

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9) David puts $4125 into an account. If he keeps the money in the account for 5 years and
now has a total of $4193.89. What is the interest rate?

4193.89= 4125e^(rt) NOTE: t = 5 given
4193.89/4125=e^(5r)
ln( 4193.89/4125 ) = ln(e^5r) = 5r
ln( 4193.89/4125 )/5 = r= 0.00331253688=.33% ans
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10) Tsonya put some money into an account paying 4.5% compounded continuously for
10 years. She now has $3567.91 in the account. How much money did she start the
account with?
  A = Pe^(rt)
3567.91 = Pe^(.045*10)
3567.91*e^(-.45) = P= 2274.99985845284 = 2275.00 Ans