SOLUTION: Moisha is saving up money for a down payment on a house. She currently has $5579, but knows she can get a loan at a lower interest rate if she can put down $6160. If she invests

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Question 1193774: Moisha is saving up money for a down payment on a house. She currently has $5579, but knows she can get a loan at a lower interest rate if she can put down $6160. If she invests the $5579 in an account that earns 5.2% annually, compounded monthly, how long will it take Moisha to accumulate the $6160? Round your answer to two decimal places, if necessary.
Found 3 solutions by Theo, MathTherapy, ikleyn:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
formula to use is f = p * (1+r) ^ n
f is the future value
p is the present value
1 + r is the growth factor per time period.
n is the number of time periods.

your interest rate per year is 5.2% / 100 = .052
your interest rate per month is .052/12 = .00433333333
your growth factor per month is 1.00433333333

the number of time periods is the number of months.

formula becomes:
6160 = 5579 * 1.00433333333 ^ n
divide both sides of the equation by 5579 to get:
6160/5579 = 1.00433333333 ^ n
take the log of both sides of the equation to get:
log(6160/5579) = log(1.00433333333 ^ n)
by one of the rules of logs, this becomes:
log(6160/5579) = n * log(1.00433333333)
divide both sides by log(1.00433333333) to get:
log(6160/5579) / log(1.00433333333) = n
solve for n to get:
n = 22.91116607 months.
confirm by replacing n in the original equation to get:
f = 5579 * 1.00433333333 ^ 22.91116607 = 6160.
your solution is that 5579 will grow to 6160 in 22.91116607 months at the rate of .00433333333 per month.

the annual rate if 5.2%.
divide that by 12 to get a monthly rate of 5.2%/12.
since 5.2% = .052, then you get a monthly rate of .052/12 = .00433333333
the growth factor per month is 1 plus that = 1.00433333333.

here are the rules of log arithmetic.
https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/
the one used in this problem is rule 3.




Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Moisha is saving up money for a down payment on a house. She currently has $5579, but knows she can get a loan at a lower interest rate if she can put down $6160. If she invests the $5579 in an account that earns 5.2% annually, compounded monthly, how long will it take Moisha to accumulate the $6160? Round your answer to two decimal places, if necessary.
SIMPLY use the following Future Value formula for a value of $1: 
         matrix%281%2C3%2C+A%2C+%22=%22%2C+P%281+%2B+i%2Fm%29%5E%28mt%29%29
     
       matrix%281%2C3%2C+12t%2C+%22=%22%2C+log+%28%281+%2B+.052%2F12%29%2C+%28880%2F797%29%29%29 ------- Converting to LOGARITHMIC form
Time, or  = 1.90926384 years ≈ 1.91 years, or approximately 1 year 10.9 months, or about 1 year and 11 months.

Answer by ikleyn(52799) About Me  (Show Source):
You can put this solution on YOUR website!
.

In this problem, after formal calculations you get the answer t= 1.909264 years, or 1 year and 10.91 months.


                But this answer contradicts the logic and is not final.


After that, you MUST round this non-integer number of months to the closest greater integer number of months
in order for the bank was in position to make the last compounding.

So the final answer is 1 year and 11 months, since the account is compound monthly.


I repeated this note about 77 times at this forum, but some tutors like @Theo ignore this necessity
to round to the closest greater integer number of compound periods, and prefer to teach incorrectly.


So I just tired to repeat it again and again.