1. ## Manipulate big equation

Can any one help to manipulate this equation so that we can get a new equation to calulate the value of N with these different variables ( P,M,Z,S and W)
9.36Log10(N+1) + {Log10(P/2.7)}/{0.4+(1094/(N+1)^5.19)} =Log10W - Z*S + 8.27 -2.32log10M

Thanks

2. This is ambigious : where should the power (^5.19) be put ? Just on the bottom part of the fraction or on the whole fraction ? And does it really have to be that twisted ? Please make the effort of presenting the question in such a way that it is relatively easy and convenient to best answer the actual question.

I'd like to make you realize how hard what you are asking could be : the variable to isolate (N) is :
- in a log, along with an addition (+1)
- outside a log along with a real power (5.19)

It might even not be possible to make N the subject using algebra only.

3. ## manipulate

Yes this is an ambigious equation, N + 1 has power 5.19 and also it is inside log10 . I think this equation is not easy to solve.
In my opinion it can only be done by recurrsion (putting N terms in one side and putting the values of different variables on the other side and checking for which values is N satisfying the equation.)

4. Could you please edit your first post then ? And eventually put it in LaTeX.
In my opinion it can only be done by recurrsion (putting N terms in one side and putting the values of different variables on the other side and checking for which values is N satisfying the equation.)
Hmm, this is not what I would call recursion. However, what I mean is that it may not even be possible to isolate all N terms on one side, you might fall into a situation where trying to move one N term on a side will make the other go on the other side. And algebra is useless in this situation. So either you :

- try to see what can be done with very, very advanced mathematics (believe me, five variables, you are going to have a lot of fun)
- need to put up with it and try to find a solution (you are gonna have fun too ...)

Here, I put it in LaTeX for you, is that right ?

$9.36 \log_{10}{(N+1)} + \log_{10}{\left ( \frac{\frac{P}{2.7}}{0.4+\frac{1094}{(N+1)^{5.19}} } \right )} = \log_{10}{(W)} - Z \times S + 8.27 -2.32 \log_{10}{(M)}$

(my eyes !)

Maybe someone will be courageous enough to give it a shot, I would've helped you but I can't stay anymore now

5. I am new to the forum world and especially with LaTeX. One of the error in your post is the log 10 is only with( P/2.7) in the left side of the equation not with the denominator.
9.36Log10(N+1) + (Log10(P/2.7))/(0.4+1094/(N+1)^5.19) =Log10W - Z*S + 8.27 -2.32log10M

Thanks Ray

Originally Posted by Bacterius
Could you please edit your first post then ? And eventually put it in LaTeX.

Hmm, this is not what I would call recursion. However, what I mean is that it may not even be possible to isolate all N terms on one side, you might fall into a situation where trying to move one N term on a side will make the other go on the other side. And algebra is useless in this situation. So either you :

- try to see what can be done with very, very advanced mathematics (believe me, five variables, you are going to have a lot of fun)
- need to put up with it and try to find a solution (you are gonna have fun too ...)

Here, I put it in LaTeX for you, is that right ?

$9.36 \log_{10}{(N+1)} + {\left ( \frac{\log_{10}{\frac{P}{2.7}}{0.4+\frac{1094}{(N+ 1)^{5.19}}} \right )} = \log_{10}{(W)} - Z \times S + 8.27 -2.32 \log_{10}{(M)}$

(my eyes !)

Maybe someone will be courageous enough to give it a shot, I would've helped you but I can't stay anymore now

6. But you just told me the power was inside the logarithm ? And like this ?

$9.36 \log_{10}{(N+1)} + \log_{10}{\left ( \frac{P}{2.7} \right ) } \times \left ( 0.4 + \frac{1094}{(N+1)^{5.19}} \right ) = \log_{10}{(W)} - Z$ $S + 8.27 -2.32 \log_{10}{(M)}$

(sorry I had to cut the equation in half, image too big)

It *might* be solvable, I would recommend putting everything that doesn't involve N to the right, so that you are left with the big brackets and the log of N + 1. Then take the log of 9.36, replace it with another variable, and combine the logs. Then exponentiate and see how you can work it out (it *might* work, I did not try).

7. Power is just above (N+1) and this term is in the denominator ( below 1094)........i.e:........1094/((N+1)^5.19)

Originally Posted by Bacterius
But you just told me the power was inside the logarithm ? And like this ?

$9.36 \log_{10}{(N+1)} + \log_{10}{\left ( \frac{P}{2.7} \right ) } \left ( 0.4 + \frac{1094}{(N+1)^{5.19}} \right ) = \log_{10}{(W)} - Z$ $S + 8.27 -2.32 \log_{10}{(M)}$

(sorry I had to cut the equation in half, image too big)

8. Well I think I got it right then.

9. Log10(P/2.7) is the numerator and {0.4 + (1094/.....)} is in the denominator
Thanks again Ray

Power is just above (N+1) and this term is in the denominator ( below 1094)........i.e:........1094/((N+1)^5.19)
Originally Posted by Bacterius
Well I think I got it right then.

10. Oh, right !

$9.36 \log_{10}{(N+1)} + \frac{\log_{10}{\left ( \frac{P}{2.7} \right ) }}{0.4 + \frac{1094}{(N+1)^{5.19}}} = \log_{10}{(W)} - Z$ $S + 8.27 -2.32 \log_{10}{(M)}$

Well, now we've spent a page formatting the equation properly, I believe you might get an answer to the original question in a little less than twenty pages

11. Now u got it perfect
Thanks
Originally Posted by Bacterius
Oh, right !

$9.36 \log_{10}{(N+1)} + \frac{\log_{10}{\left ( \frac{P}{2.7} \right ) }}{\left ( 0.4 + \frac{1094}{(N+1)^{5.19}} \right )} = \log_{10}{(W)} - Z$ $S + 8.27 -2.32 \log_{10}{(M)}$

Well, now we've spent a page formatting the equation properly, I believe you might get an answer to the original question in a little less than twenty pages

12. Well ... this actually increases the complexity of solving what I thought was the equation hundredfolds
I hope you have A2 paper, coz' this will be big.