Thread: How do I extract j from, B = A.(1 + j)^N + (P/j).[(1 + j)^N - 1]?

How do I extract j from, B = A.(1 + j)^N + (P/j).[(1 + j)^N - 1]?

I think it can be useful to use a math program to calculate that for you.

i don't have any. i tried online using wolframaplha but it didn't help...it's really difficult to extract j.. hope someone can show a solution...

made a bad error... sorry!

Originally Posted by prinzrainer

i don't have any. i tried online using wolframaplha but it didn't help...it's really difficult to extract j.. hope someone can show a solution...

Yes, I also tried wolphram alpha but that didn't help, but I'm not sure if this can be done without a math program.

Originally Posted by prinzrainer

How do I extract j from, B = A.(1 + j)^N + (P/j).[(1 + j)^N - 1]?

This is the formula to calculate the Future value of (usually) a savings account
with an initial Amount deposited, then regular (usually) monthly Payments (deposits).

The standard way to show is: F = A(1 + i)^n + (P / i)[(1 + i)^n - 1] , where:
F = Future value, A = initial Amount deposited, P = regular Payment,
n = Number of regular payments, i = periodic Interest rate.

As example, $1000 deposited initially, then $100 per month for 36 months at annual
rate of 6% compounded monthly ( i becomes .06/12 = .005):
F = 1000(1.005)^36 + (100 / .005)(1.005^36 - 1) = 5130.29102....

It is not possible to solve for i directly; must be done numerically.
If you google "iteration", you'll get an idea of what's involved.
It really becomes "guessing", then "homing in" until you hit paydirt!