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How would you simplify this?

After dividing 80,000 / 587, would I divide 0.0067 with all the terms in the brackets.

80,000 = 587 [ 1 - (1.0067)^(-12 t) / 0.0067 ]

To make it clear, the first part in the brackets [ ] divided by 0.0067

I suppose I have to use logarithms... ?



Thx! :D

2007-06-11 09:29:25 · 3 answers · asked by jill b 1 in Science & Mathematics Mathematics

3 answers

If you want to solve this for t, then you do need to use logarithms.

80,000 = 587 [ ( 1 - (1.0067)^(-12 t) ) / 0.0067 ]
Divide by 587:
80000/587 = 1 - (1.0067)^(-12 t) ) / 0.0067
Multiply by 0.0067:
80000*0.0067/587 = 1 - (1.0067)^(-12t)
Subtract 1 and change signs:
1 - 80000*0.0067/587 = 1.0067^(-12t)
Take logs:
log(1 - 80000*0.0067/587 ) = -12t log(1.0067)
Divide by -12 log(1.0067):
t = - log(1 - 80000*0.0067/587 ) / 12 log(1.0067)
= 30.49.

2007-06-11 09:42:46 · answer #1 · answered by Anonymous · 0 0

I’m unclear how you got this far. It looks as if it might be a kind of compound interest problem… or maybe an annuity problem… I’m guessing you have an annual interest rate of 8%, and the installments are monthly. It would help greatly to see the formula into which you plugged these values. Anyway…
80,000 = 587 [ 1 - (1.0067)^(-12 t) / 0.0067 ]

If you mean: 80,000 = 587 [ 1 – {(1.0067)^(-12 t)} / 0.0067 ], and this is the way you have it written...
Then you only divide the {(1.0067)^(-12 t)} by .0067.
Then you take 587 times that value from 587.

But if you mean 80,000 = 587 [{1 - (1.0067)^(-12 t)} / 0.0067 ]
Then you divide 1 - (1.0067)^(-12 t) by .0067.
Then you multiply that result by 587.

And, yes. you'll have to use logarithms.

2007-06-11 09:57:00 · answer #2 · answered by gugliamo00 7 · 0 0

You MUST be clearer with your brackets.

I don't know what you were going on about when you said "the first part in the brackets etc".

Why didn't you just put that down in the formula?

Brackets are so easy to use and make things so clear, there's no excuse not using them!

2007-06-12 01:07:37 · answer #3 · answered by tuthutop 2 · 0 0

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