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morgaine300 · Sep 4, 2009, 09:56 PM

Hi all,

Those of you who know me know I'm not doing homework, so save my speech. :p

Actually, I need this for calculating an accounting thing.

I know how to solve this:

$6 = 1.05^n$

using the LN key. I do not know anything about how this works because I never learned log stuff. I just have that one memorized. (Yes, morgaine actually does plug n chug occasionally. Don't tell anyone.)

Therefore, I have no clue how to apply it to this:

$-6 = 1.05^{-n}$ .

When I use my financial calculator for the accounting problem, I'm getting -36.72, which is simply the negative of:

$\frac{6\ ln}{1.05\ ln}$ .

... and that doesn't check back out. So anyone got an easy way to attack this?

Thanks.

morgaine300 · Sep 4, 2009, 10:04 PM

Oh well poo.

Is that example even solvable? I just went to Excel to try to solve it the long way, just to get the answer so I could play around with this, and discovered the accounting problem given is impossible to begin with. So that answers that.

However, I'd still like to know how to solve for the negative exponent, cause this is certainly not the first time I've run across this, nor will it be the last.

(For anyone who cares, I've been trying to work with this):

$PV=Pmt\,\left(\frac{1-(1+i)^{-n}}i\right)$

I can get down to:

$a= b^{-n}$ and don't know how to do the negative.

(And oops, I forgot there's another math forum that isn't homework.)

ArcSine · Sep 5, 2009, 05:52 AM

Neg exponents are defined such that $a^{-b}$ is just another way of writing $\frac{1}{a^b}$ (undefined, of course, for a = 0). So that PV-of-an-annuity you're working with is just an alternate rendering of the familiar

$\frac{1-\frac{1}{(1+i)^n}}{i}$

With that, you can see where $-6=1.05^{-n}$ crashes and burns.

$1.05^{-n}\ =\ \frac{1}{1.05^n}$ , and there is no n which can turn that fraction into a negative number. As n ranges across $(-\infty,\ \infty)$ that fraction will always stay north of zero.

morgaine300 · Sep 5, 2009, 05:10 PM

So that PV-of-an-annuity you're working with is just an alternate rendering of the familiar

$\frac{1-\frac{1}{(1+i)^n}}{i}$

No, this is the 'alternative' to my familiar equation. :D (And mine's easier to do on the calculator.)

$1.05^{-n}\ =\ \frac{1}{1.05^n}$ , and there is no n which can turn that fraction into a negative number. As n ranges across $(-\infty,\ \infty)$ that fraction will always stay north of zero.

Yup, I see what you're saying. I'm right now trying to figure out if that'll ever end up positive... hang on a sec...

I'm back. :p

OK... I see why it came out negative and why it shouldn't have. (You may have seen that post where the payment wouldn't work.) So I just did it with a different payment of 10,000 and it's coming down to:

$.5 = 1.05^{-n}$

I would assume that would normally be a positive on the left. As long as the payment amount covers the first interest, it has to be solvable. So the question remains: how to solve for the negative exponent? (Yes, I know it's $\frac{1}{1.05^{n}}$ but I don't know how to solve that.)

galactus · Sep 5, 2009, 05:48 PM

$.5=\frac{1}{1.05^{n}}$

$ln(.5)=-nln(1.05)$

$n=-\frac{ln(.5)}{ln(1.05)} = \frac{ln(2)}{ln(\frac{21}{20})} \approx {14.207}$

You're certainly correct about the -6 and the negative exponent.

No real solution in that.

It does have a solution though, if we were to solve it in the complex realm we would have:

$n=\frac{-ln(6)}{ln(\frac{21}{20})}+\frac{\pi}{ln(\frac{21}{ 20})}i$

morgaine300 · Sep 5, 2009, 07:48 PM

$.5=\frac{1}{1.05^{n}}$

$ln(.5)=-nln(1.05)$

$n=-\frac{ln(.5)}{ln(1.05)} = \frac{ln(2)}{ln(\frac{21}{20})} \approx {14.207}$

In other words, it's just the negative of the answer to $.5=1.05^{n}$ ? I mean, that's it?

I don't understand logs, but I can certainly memorize that for the sake of my accounting junk. :)

morgaine300 · Sep 5, 2009, 08:14 PM

Um... does anyone want to tackle explaining this one?

$FV=Pmt\,\left(\frac{(1+i)^n-1}i\right)$

$PV=Pmt\,\left(\frac{1-(1+i)^{-n}}i\right)$

How to solve for the i. (That's interest, not an imaginary number. I'm sure anyone with a loan wishes it were imaginary. :p)

I can get as far as dividing both sides by the payment. ;)

I have a financial calculator (though I generally do the algebra instead of using it), but even it gives me "no solution" when solving for i. So I basically am stuck with no way to solve for these.

I have a feeling the solution to this will make about as much sense as when someone tried to show me how the quadratic formula was derived - we have an i under an exponent and one not. So I might not get it.

Let's stick some numbers in there. They should come out to around 5%.

$100,000=1000\,\left(\frac{(1+i)^{37}-1}i\right)$

$100,000=10,000\,\left(\frac{1-(1+i)^{-14}}i\right)$

And thank you for putting up with me.

galactus · Sep 6, 2009, 06:51 AM

That is a booger to solve for i.

To works out to be the roots of $Vx-mt+mt(1+x)^{-n}=0$

But, here is a way we can solve a similar formula for i.

Take the formula for compounding interest.

$A=P(1+\frac{i}{n})^{nt}$

$\frac{A}{P}=(1+\frac{i}{n})^{nt}$

$ln(\frac{A}{P})=\underbrace{nt\cdot ln(1+\frac{i}{n})}_{\text{log law, ln(a^b)=bln(a)}}$

$e^{\frac{ln(\frac{A}{P})}{nt}}=1+\frac{i}{n}$

$n\cdot \left(e^{\frac{ln(\frac{A}{P})}{nt}}-1\right)=i$

or equivalently:

$i=n\left[(\frac{A}{P})^{\frac{1}{nt}}-1\right]$

For some friendly advice, learn the log laws. They are very prevalent in the math word and are woth knowing. They are in any algebra book. I listed one above.

For instance, $log(ab)=log(a)+log(b)$

$log(\frac{a}{b})=log(a)-log(b)$

$log(a^{b})=blog(a)$

$log_{a}b=c \Rightarrow a^{c}=b$

ArcSine · Sep 6, 2009, 07:29 AM

Yep, solving for the interest rate i in Morgaine's annuity problem is a job for a lightning-fast trial-and-error application, like Excel's Goal Seek. I'd imagine a calculator might just crash and run up the "no solution" white flag if the number of iterations it had to go through is too high.

The present-value-of-an-annuity equation, when fully written out, is just a polynomial, with the number of terms = the number of payments (360, e.g. for a typical 30-year fixed rate home mortgage).

The reason it can be reduced to that nifty shortcut formula is that the poly forms a Geometric Series. It's easy to solve for the initial amount, or the annuity payment. But solving for i means, as Galactus said, finding the roots. And with a (e.g.) 360-term poly, well... (although only one of those roots will "make sense" in the physical context of a financial annuity).

(Side note, Morgaine: You've probably heard, in the classic "which is superior, NPV or IRR?" argument, that one of the flaws of IRR is that it can have multiple solutions? That multi-root situation is the culprit.)

Cheers, all

morgaine300 · Sep 6, 2009, 06:03 PM

You both get greenies just for trying to help. Sounds like trying to do the annuity for I is just not going to work, at least not for me. I half suspected as much. I can actually solve it by trial & error to 2 decimals in a reasonable amount of time.

galactus, I already can solve the compounded ones for nearly anything. Your n is not the same as mine - yours is representing the number of compounding periods in a year, and I've never seen occasion to have to solve for that. My n is what you have as nt, and that's pretty easy.

And ArcSine, I don't know how to do IRR. I learned it in finance but not something I remember. So no I'm not familiar with that argument. :) (I'm not a finance person, remember? And I can only do easy NPV's.)

Well, that was a fun, huh?

morgaine300 · Sep 6, 2009, 06:05 PM

I lied. ArcSine doesn't get any green coins today cause I haven't given enough others away yet. :-)