pre calc question

Question

A savings account pays interest at a annual percentage rate (APR) of 3.4%, compounded monthly.

A customer opens a savings account with an initial deposit of $1000. Assuming no further deposits or withdrawals, after how many years will the value of the account reach $1400?

Round your answer to three decimal places.

Answer: The value will reach $1400 after  years.

Answer 1

general formula is:

A =P(1+r/n)^nt

where P = starting amount= $1000

A = ending Amount = $1400

r = APR= annual percentage rate of interest= 3.4% = .034

n = number of compounding periods per year

t = time in years = 9.910

plug in the numbers, and solve for t

1400=1000(1+.034/12)^12t

divide by 1000

1400/1000

= 7/5 = (1.028333...)^12t

take logs to the base 1.0028333..., and divide by 12

t = [log_1.0028333...1.4]_/12

_{= ln(1.4)/12ln1.0028333...}

_{= .336472236/1.002829327007}

_{= (.336472236)(29.45341176)}

_{= 9.910255331 years = about 9.910 years}

_{= about 9 years, 332 days, 11 hours, 4 minutes, 47 seconds, rounded off to nearest second}

Answer 2

This question is asking us to apply our knowledge of the equation for compound interest of the form A(t)=P(1+(r/n)^(nt)

Lets break down this formula A(t)=P(1+(r/n)^(nt)

1. t, our independent variable is defined as the number of years from the point in time where interest begins to accrue. This is the value we are looking for
2. A, our dependent variable is the amount of money in the savings account after t years, the problem tells us this is 1400
3. P, is the principle or the original amount of money placed into the account at t=0, the problem tells us this is 1000
4. r is the nominal annual percentage rate or APR, the problem tells us this is 3.4%, or 0.034
5. n is the number of times per year we compound our interest, the problem tells us this is twelve since we compound monthly and there are 12 months in a year

Next we solve our equation for t in order to get an equation we can substitute our values into in order to get the answer.

A=P(1+(r/n)^(nt)

(A/P)=(1+(r/n)^(nt)

log(A/P)=log(1+(r/n)^(nt) ( we take a logarithm of both sides as a strategy to remove the exponent

log(A/P)=ntlog(1+(r/n)) (a property of logarithms log(a)^b=blog(a)). Now t is no longer in the exponent

log(A/P)/(nlog(1+(r/n))=t

Now we substitute the value we were given in the problem

t=log(1400/1000)/(12log(1+(0.034/12))

t=log(1.4)/(12log(1.00283))

t=(0.1461)/(12(0.0012288))

t=9.908

The value will reach $1400 after 9.908 years