30000 = x [1 + (.075 / 12)]^(15 * 12)
log(30000) = log(x) + 180 log(1.0625)
to establish a college fun, what lump sum must be deposited in order to have $30,000.00 in the fund at the end 0f 15 years? Assume a 7.5% interest rate, compounded monthly
3 answers
log(30000) = log(x) + 180 log(1.0625)
should read: log(30000) = log(x) + 180 log(1.00625)
or
from the first line
300000 = x(1.00625)^180
x = 30000/1.00625^180 = 9773.73
The log equations yields the same result.
should read: log(30000) = log(x) + 180 log(1.00625)
or
from the first line
300000 = x(1.00625)^180
x = 30000/1.00625^180 = 9773.73
The log equations yields the same result.
oops...decimal place
sorry about that
sorry about that
2 answers