1. Solve for n

This is a sum from example exercises I might get on my entrance exam - and I've no idea how to solve it! You're supposed to rearrange the terms in terms of n.

I can take the (q-1) to the left, take the q^n to the right, take r to the left as well... and that's when the problems begin.

I know logs are involved in it somehow...
Should I completely ignore the little 0 next to the R? Mathematically speaking, are R and r different variables?

2. Re: Solve for n

Originally Posted by Aluminium

This is a sum from example exercises I might get on my entrance exam - and I've no idea how to solve it! You're supposed to rearrange the terms in terms of n.

I can take the (q-1) to the left, take the q^n to the right, take r to the left as well... and that's when the problems begin.
What problems? Can you show the steps you have worked out? We will help you out.

I know logs are involved in it somehow...
Should I completely ignore the little 0 next to the R? Mathematically speaking, are R and r different variables?
Yes R and r are different variables.

3. Re: Solve for n

\displaystyle \displaystyle \begin{align*} R_0\,q^n &= \frac{r(q^n - 1)}{q - 1} \\ R_0(q - 1)q^n &= r(q^n - 1) \\ R_0(q - 1)q^n &= r\,q^n - r \\ R_0(q - 1)q^n - r\,q^n &= -r \\ [R_0(q - 1) - r]q^n &= -r \\ q^n &= -\frac{r}{R_0(q - 1) - r} \\ q^n &= \frac{r}{r - R_0(q - 1)} \\ \ln{\left(q^n\right)} &= \ln\left[\frac{r}{r - R_0(q - 1)}\right]\\ n\ln{q} &= \ln\left[\frac{r}{r - R_0(q - 1)}\right] \\ n &= \frac{\ln{\left[\frac{r}{r - R_0(q-1)}\right]}}{\ln{q}} \end{align*}

4. Re: Solve for n

"Prove it", it's supposed to be in terms of n, not q.

This is what I have so far:

(I can't find how to do the little 0 next to the R)

To get that dreaded n, I'm supposed to isolate it to one side. I'm at loss...

5. Re: Solve for n

Originally Posted by Aluminium

This is a sum from example exercises I might get on my entrance exam - and I've no idea how to solve it! You're supposed to rearrange the terms in terms of n.

I can take the (q-1) to the left, take the q^n to the right, take r to the left as well... and that's when the problems begin.

I know logs are involved in it somehow...
Should I completely ignore the little 0 next to the R? Mathematically speaking, are R and r different variables?
If n as function of the other variables is required is...

$\displaystyle n = - \frac{\ln \{1- \frac{R_{0}}{r}\ (q-1)\}}{\ln q}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. Re: Solve for n

Originally Posted by Aluminium
"Prove it", it's supposed to be in terms of n, not q.

This is what I have so far:

(I can't find how to do the little 0 next to the R)

To get that dreaded n, I'm supposed to isolate it to one side. I'm at loss...
Just divide from there..

$\displaystyle \frac{R_{0}(q-1)}{r}=\frac{q^n-1}{q^n}=\frac{q^n}{q^n}-\frac{1}{q^n}=1-\frac{1}{q^n}$

You now have a "single" n.

$\displaystyle \Rightarrow\frac{1}{q^n}=1-\frac{R_{0}(q-1)}{r}=\frac{r}{r}-\frac{R_{0}(q-1)}{r}=\frac{r-R_{0}(q-1)}{r}$

$\displaystyle \Rightarrow\ q^n=\frac{r}{r-R_{0}(q-1)}$

Now finish by taking logs of both sides and use the power law.

I'm not sure that's what you want,
because you say "rearrange in terms of n".
That means some term = f(n).

Do you want n=?
which is "n in terms of the others".

7. Re: Solve for n

Changing your R,q,r to a,b,c: ab^n = c(b^n - 1) / (b-1)

Simplify:
ab^n(b - 1) = cb^n - c
Divide by b^n:
a(b - 1) = c - c/b^n
Re-arrange:
b^n = c / [c - a(b - 1)]
Wrap up:
n = LOG{c / [c - a(b - 1)]} / LOG(b)

8. Re: Solve for n

Originally Posted by Aluminium
"Prove it", it's supposed to be in terms of n, not q.

This is what I have so far:

(I can't find how to do the little 0 next to the R)

To get that dreaded n, I'm supposed to isolate it to one side. I'm at loss...
Oops - Edited

9. Re: Solve for n

Originally Posted by Aluminium
"Prove it", it's supposed to be in terms of n, not q.
You titled this thread "solve for n" which is exactly what Prove it did. My German is a bit rusty but I would have interpreted "nach n auf" as "for n"- especially since there are
4 parameters and you can 'solve' for one of them 'in terms of' the other three.

This is what I have so far:

(I can't find how to do the little 0 next to the R)

To get that dreaded n, I'm supposed to isolate it to one side. I'm at loss...

10. Re: Solve for n

Oh, then my terminology is incorrect, sorry about that.
Thank you for all the help, everyone. I'm going to try doing it now and cross-check with your answers.

Are these sort of equations termed anything specific? I'd like to look for them online.