SOLUTION: 1)Find the time required for an investment of 7,000 dollars to grow to 11,000 dollars at an interest rate of 6% per year, compounded monthly. Give your answer accurate to 2 decimal

Question 1161154: 1)Find the time required for an investment of 7,000 dollars to grow to 11,000 dollars at an interest rate of 6% per year, compounded monthly. Give your answer accurate to 2 decimal places.
2) Joy invests $19,000 at 14% simple interest for 1 year. How much is in the account at the end of the 1 year period?

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
1)
f = p*(1+r)^n
f is the future value
p is the present value
r is the interest rate per time period
n is the number of time periods
your equation is:
11000 = 7000 * (1+.06/12)^n
your time period is in months.
the interest rate per month is .06/12
the number of months is n.
divide both sides of that equation by 7000 to get:
11000/7000 = (1+.06/12)^n
take the log of both sides of this equation to get:
log(11000/7000) = log((1+.06/12)^n)
since log(x^a) = a*log(x), this becomes:
log(11000/7000) = n*log((1+.06/12)
divide both sides of this equation by log(1+.06/12)) and solve for n to get:
n = log(11000/7000)/log(1+.06/12) = 90.62282945.
this says that 7000 will become 11000 in 90.62282945 months.
to confirm, replace n in the original equation to get:
f = 7000*(1+.06/12)^90.6228i2945.
solve for f to get:
f = 11000
this confirms the solution is correct.
90.622812945 months divided by 12 is equivalent to 7.551902454 years.
2)
19000 * 1.14 = 21660.
that's what's in the account at the end of the year.
simple interest formula is f = p*p+r*n which can be simplified to f = p*(1+r*n)
in this problem:
p = 19000
r=.14 per year
n = 1 year.
formula becomes f = 19000*(1+.14) = 19000 * 1.14 = 21660.

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