AuthorTopic: Mathematics problem
happy123
@2014-02-02 06:31:05
Hi

Do you know how to solve this problem:

1.0605 = (1+(0.0589/m))power m

m = ?

Thank you.
aLEx78
@2014-02-02 17:37:08
m=12

means a monthly compounding frequency rate
happy123
@2014-02-03 12:47:58
Hi

Is it possible for you to show me the working instead of just the answer.

Thank you.
happy123
@2014-02-03 13:05:41
Another mathematics problem is:

100,000= 35,000 (1 + (5%/12))power 12N

N=?

Thank you.

If there is an alpha in the power, I do not know how to turn the power into normal equation to get to the answer.
muna
@2014-04-30 18:37:12
Hi Mr Happy!

In doing both level I and level II, I have not come across such a problem.
Your first question puzzled me for a while until I realised, as alex poited out, that 'm' refers to the number of months. (just means the rate calculated is compounded monthly)

Theoretically, the equation can be reduced to a quadratic - i have done it but takes time to explain. Rest assured you won't be required to do this, but I suppose you misinterpreted the question.

Your last question leads me to beleive the following simple algrbra will cover quite a few of your problems...
If K = Y^p, where ^ = power of.
then Y = K^(1/p)

eg. 9 = 3^2, 3 = 9^(1/2).

Cheers mate.
Muna.
4 = 2^2