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Hoddi · Mar 2, 2010, 05:44 AM

Can anyone please tell me what part of maths this falls under and what I need to work it out?

Unknown008 · Mar 2, 2010, 09:01 AM

I think this falls in the category of logarithms.

You need to replace the variables by the given value. Then, you'll have to proceed to a change from index form to the logarithm form.

Can you do it?

morgaine300 · Mar 3, 2010, 02:56 AM

You need the LN key.

Actually, it's A = P(1 + R)^n. I don't know what that 0,01 is about. But you're not multiplying anything by the R. There's actually different ways to write that equation, but yours hasn't given any kind of compounding so just make the parenthesis 1 + R.

Divide both sides by P. Keep in mind the parenthesis and exponent must stay together, so you've got to get rid of the P.

Once you have it to a $A = B^n$ format, then do:

$\frac{ln\ A}{ln\ B}\ =\ N$

Unknown008 · Mar 3, 2010, 07:01 AM

Well, you can use the log key also, this won't change a thing.

However, if the 0.01 was indeed in the problem, there would still be a solution for the equation, right? I wonder why you say the 0.01 shouldn't be there :confused:

Hoddi · Mar 3, 2010, 08:42 AM

It is 0,01 in the question.. it does not make this solution any easyer.

Unknown008 · Mar 3, 2010, 09:28 AM

Put in the values given.

$A=P(1+0.01R)^N$

$(575)=(90)(1+0.01(0.17))^N$

Ok? Well, divide both sides by 90 and simplify the number in brackets.

$\frac{575}{90}=\frac{\cancel{90}(1+0.01(0.17))^N}{ \cancel{90}}$

$6.3888...=(1.0017)^N$

Insert log or ln on both sides.

$ln\ (6.3888...)=ln\ (1.0017)^N$

When you have ln [something]^n, that is equal to n ln [something].

So, you put can put it to:

$ln\ (6.3888...)=N\ ln\ (1.0017)$

Divide both sides by ln (1.0017):

$\frac{ln\ (6.3888...)}{ln\ (1.0017)}=\frac{N\ \cancel{ln\ (1.0017)}}{\cancel{ln\ (1.0017)}}$

$\frac{ln\ (6.3888...)}{ln\ (1.0017)}=N$

Now grab a calculator and work out the value of N.

morgaine300 · Mar 5, 2010, 07:46 PM

However, if the 0.01 was indeed in the problem, there would still be a solution for the equation, right? I wonder why you say the 0.01 shouldn't be there

Because A stands for amount (future amount) and P is principal, R is rate and N is number of periods. It's a typical future value equation. Having .01 in it doesn't make any sense. It means you're taking a year's worth of interest and taking 1/100th of it. So that means it's compounding every 3.65 days. R could conceivably be used for the rate per compounding period, but then you're still getting 1/100th of that, which still doesn't have much meaning.

So while mathematically it can still be solved, it doesn't make any sense within the context of what that equation is.

Unknown008 · Mar 6, 2010, 08:23 AM

Oh, well, me not doing accounts have some limits as to those. Thanks for clearing things up :)

morgaine300 · Mar 9, 2010, 02:33 AM

No problem. :-)