Solve for n using logarithm

[getoutofmylaboratory]

A = (Pr)/[1-(1+r)^-n] (Solve for n, using logarithms with appropriate bases.

Here's where I've come to so far, but not sure if on the right track:

A = (Pr)/[1-(1+r)^-n] =>
[1 - (1 + r)^-n]A = Pr => --(got rid of the fraction to hopefully make it easier)
(1 - 1 - r^-n)A = Pr => -- (distributed negative)
(-r^-n)A = Pr => -- (1 -1 cancels out 1)
-Ar^-n = Pr => -- (distribute A inside parenthesis)

Okay, here's the part where I think the log comes in:

log(-Ar^-n) = Pr =>
n log Ar = -Pr => (multiply both sides by -1)


Here's where I need help (but of course, point out any other errors I've made). I don't think that log is quite right.

Thanks,

 

[getoutofmylaboratory]

******** NO :evil:
All you can do here is:
Pr = A - A(1 + r)^(-n)

Try again...
Remember this li'l rule: a^(-p) = 1 / a^p

Was this YOU:
http://www.youtube.com/watch?v=l2PoSljk8cE

Ok, so I've been playing around with this.

This is as far as I can figure out. I can't think of a way to get (1 + r)^n by itself so that I can take the natural log to get the answer.

Pr = A - A(1 + r)^-n =>
Pr = A - A * 1/(1 + r)^n =>
Pr = A - A/(1 + r)^n

Now what? I could get rid of the fraction again...but I'm not sure that leaves me any better off than before?
Pr(1 + r)^n = A - A(1 + r)^n

Or...what if I start moving the A around?

Pr - A = - A/(1 + r)^n => (subtract A from both sides)
A - Pr = A/(1 + r)^n => (multiple both sides by -1)

Am I getting close?

 

[JeffM]

Ok, so I've been playing around with this.

This is as far as I can figure out. I can't think of a way to get (1 + r)^n by itself so that I can take the natural log to get the answer.

Pr = A - A(1 + r)^-n =>
Pr = A - A * 1/(1 + r)^n =>
Pr = A - A/(1 + r)^n

Now what? I could get rid of the fraction again...but I'm not sure that leaves me any better off than before?
Pr(1 + r)^n = A - A(1 + r)^n

Or...what if I start moving the A around?

Pr - A = - A/(1 + r)^n => (subtract A from both sides)
A - Pr = A/(1 + r)^n => (multiple both sides by -1)

Am I getting close?

You want to solve for n. Remember way back in beginning algebra, you were told to isolate the variable to be found?

This is NO different. Because n is an exponent you want to get (1 + r)^n alone on one side of the equation. Do you see why?

\(\displaystyle A = \dfrac{Pr}{1 - (1 + r)^{-n}} \implies A\{1 - (1 + r)^{-n}\} = Pr \implies 1 - (1 + r)^{-n} = \dfrac{Pr}{A} \implies\)

\(\displaystyle -(1 + r)^{-n} = \dfrac{Pr}{A} - 1 = \dfrac{Pr - A}{A} \implies (1 + r)^n = \dfrac{A - Pr}{A}.\)

What next?

 

[lookagain]

Ok, so I've been playing around with this.

This is as far as I can figure out. I can't think of a way to get (1 + r)^n by itself so that I can take the natural log to get the answer.

Pr = A - A(1 + r)^-n =>

Pr = A - A * 1/(1 + r)^n =>

Pr = A - A/(1 + r)^n *

Now what? I could get rid of the fraction again...but I'm not sure that leaves me any better off than before?

Pr(1 + r)^n = A - A(1 + r)^n \(\displaystyle \ \ \ \)The signs on the right-hand side of the equation are opposite of what they should
be if you multiply each side of he equation in *. Also, you have complicated it more by having n in two places now.

Or...what if I start moving the A around?

Pr - A = - A/(1 + r)^n => (subtract A from both sides)

A - Pr = A/(1 + r)^n => (multiply both sides by -1)

Am I getting close? \(\displaystyle \ \ \ \)closer

\(\displaystyle A - Pr \ = \ \dfrac{A}{(1 + r)^n} \ \implies\)


\(\displaystyle \dfrac{A - Pr}{1} \ = \ \dfrac{A}{(1 + r)^n} \ \implies\)


You could invert each fraction:


\(\displaystyle \dfrac{1}{A - Pr} \ = \ \dfrac{(1 + r)^n}{A} \ \implies\)



If you choose this route, what would you do next?

 

[getoutofmylaboratory]

You want to solve for n. Remember way back in beginning algebra, you were told to isolate the variable to be found?

This is NO different. Because n is an exponent you want to get (1 + r)^n alone on one side of the equation. Do you see why?

\(\displaystyle A = \dfrac{Pr}{1 - (1 + r)^{-n}} \implies A\{1 - (1 + r)^{-n}\} = Pr \implies 1 - (1 + r)^{-n} = \dfrac{Pr}{A} \implies\)

\(\displaystyle -(1 + r)^{-n} = \dfrac{Pr}{A} - 1 = \dfrac{Pr - A}{A} \implies (1 + r)^n = \dfrac{A - Pr}{A}.\)

What next?

Yes, I've been trying to isolate the variable, just needed some guidance to get there. Not sure why I didn't think to divide by A. So, I think I understand what we have so far:

A = Pr/[1 - (1 + r)^-n =>...................original equation
A[1 - (1 + r)^-n] = Pr =>..................multiply each side by denominator to eliminate fraction
1-(1 + r)^-n = Pr/A =>...................divide each side by A
-(1 + r)^-n = Pr/A - 1 = (Pr - A)/A.................. subtract 1 from each side. Pr/A - 1 is the same as (Pr - A)/A
(1 + r)^n = (A - Pr)/A..................multiply each side by -1

So, now that we have (1 + r)^n, it looks like I should be able to just take the natural log, right?
ln (1 + r)^n = ln (A - Pr)/A
n = ln [(A - Pr)/A]

This, to me, looks like it's as far as it can go.

 

[JeffM]

Yes, I've been trying to isolate the variable, just needed some guidance to get there. Not sure why I didn't think to divide by A. So, I think I understand what we have so far:

A = Pr/[1 - (1 + r)^-n =>...................original equation
A[1 - (1 + r)^-n] = Pr =>..................multiply each side by denominator to eliminate fraction
1-(1 + r)^-n = Pr/A =>...................divide each side by A
-(1 + r)^-n = Pr/A - 1 = (Pr - A)/A.................. subtract 1 from each side. Pr/A - 1 is the same as (Pr - A)/A
(1 + r)^n = (A - Pr)/A..................multiply each side by -1

So, now that we have (1 + r)^n, it looks like I should be able to just take the natural log, right?
ln (1 + r)^n = ln (A - Pr)/A
n = ln [(A - Pr)/A]

This, to me, looks like it's as far as it can go.

No, no. Be careful with those logs.

\(\displaystyle Given\ a > 0\ a = b \iff log_c(a) = log_c(b).\) This by the way was why we had to multiply through by - 1.

\(\displaystyle (1 + r)^n = \dfrac{A - Pr}{A} \implies ln\left\{(1 + r)^n\right\} = ln\left(\dfrac{A - Pr}{A}\right) \implies\)

\(\displaystyle n * ln(1 + r) = \ln(A - Pr) - ln(A) \implies n = \dfrac{ln(A - Pr) - ln(A)}{ln(1 + r)}.\)

 

[getoutofmylaboratory]

No, no. Be careful with those logs.

\(\displaystyle Given\ a > 0\ a = b \iff log_c(a) = log_c(b).\) This by the way was why we had to multiply through by - 1.

\(\displaystyle (1 + r)^n = \dfrac{A - Pr}{A} \implies ln\left\{(1 + r)^n\right\} = ln\left(\dfrac{A - Pr}{A}\right) \implies\)

\(\displaystyle n * ln(1 + r) = \ln(A - Pr) - ln(A) \implies n = \dfrac{ln(A - Pr) - ln(A)}{ln(1 + r)}.\)

I can definitely see why remembering the rules are very important. At first I was confused how you came up with ln (A - Pr) - ln (A). But then I realized that ln (a/b) = ln (a) - ln (b). Likewise, ln (ab) = ln (a) + ln (b).

Thanks everyone for your help. I'm going to go digest this some more.