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April 23, 2014

April 23, 2014

Posted by **lynda** on Thursday, November 14, 2013 at 9:14pm.

- interest math -
**Reiny**, Thursday, November 14, 2013 at 9:23pmI am assuming you are using the standard compound interest method, simple interest is seldom used for periods longer than a year

8000(1.0525)^n = 10100

1.0525^n = 1.2625

n log 1.0525 = log 1.2625

n = 4.5554 years , assuming the rate is 5.25 per annum compounded annually

at continuous compounding

8000 e^(.0525n) = 10100

e^(.0525n) = 1.2625

take ln of both sides

.0525n = ln 1.2625

n = 4.4399 years

btw:

at simple interest

8000 + 8000(.0525)n = 10100

n = 5 years

none of the 3 ways I used produces your answer, I have no idea how you got that.

- interest math -
**steeve**, Thursday, November 14, 2013 at 9:26pmI belive its 5 years.( asuming the interest rate is anual)

8000x .0525 = 420

(Check[8000/420*5=100])

so 10100-8000 = 2100

Finally 2100/420 = 5

Hope that's right.

- interest math -
**steeve**, Thursday, November 14, 2013 at 9:26pmI belive its 5 years.( asuming the interest rate is anual)

8000x .0525 = 420

(Check[8000/420*5=100])

so 10100-8000 = 2100

Finally 2100/420 = 5

Hope that's right.

- interest math -
**steeve**, Thursday, November 14, 2013 at 9:34pmI belive its 5 years.( asuming the interest rate is anual)

8000x .0525 = 420

(Check[8000/420*5=100])

so 10100-8000 = 2100

Finally 2100/420 = 5

Hope that's right.

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