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Rearrange for X - cant seem to make any sense of this problem

Please see attached formula picture. ee(P), ee(S), c and E are variables

I really want to rearrange to ee(P), or ee(S).

I want to make seperate plots of ee(P) and ee(S), where 0 < c < 1, at different values of E from 3 to 200.

However how hard I try to rearrange, I cant seem to get any sense of it.

What I want are two equations of;

ee(P) = a function of C at a specific set E

ee(S) = a function of C at a specific set E

Please help!

Re: Rearrange for X - cant seem to make any sense of this problem

I'm gonna take a stab in the dark here and guess what you're trying to do... (I'm sick and my brain isn't 100% working)

So you want to rearrange this: $\displaystyle E = \dfrac{\ln[(1-c)(1-ee(S))]}{\ln[(1-c)(1+ee(S))]}$ and solve for ee(S)??

From the law of logarithms, it can be re-written as: $\displaystyle E = \ln[(1-c)(1-ee(S)) - (1-c)(1+ee(S))]$

Then get rid of the logarithm by raising both sides by exp: $\displaystyle e^{E} = (1-c)(1-ee(S)) - (1-c)(1+ee(S))$

Then just expand and solve for ee(S). I think you can carry on from here.

Re: Rearrange for X - cant seem to make any sense of this problem

But what is ee(P)?

-Dan

Edit: Oh, Educated has a typo in there.

@vaffel: When you solve this for, say ee(S), you will have to do it in terms of ee(P). Is this a possible problem?

@Educated: $\displaystyle \frac{ln(a)}{ln(b)} \neq ln(a - b)$. The rule you are trying to think of is $\displaystyle ln(a) - ln(b) = ln \left ( \frac{a}{b} \right )$

- Dan

Re: Rearrange for X - cant seem to make any sense of this problem

I thought it would be easier to state my error here, rather than in my previous post as it is getting cluttered.

I now understand the ee(S) and ee(P) thing. Sorry for the goof!

-Dan

Re: Rearrange for X - cant seem to make any sense of this problem

Me again. :)

These equations can't be solved exactly for ee(S) or ee(P). A quick demonstration with a similar problem will show you why. Solve

$\displaystyle E = \frac{ln(a)}{ln(b)}$

$\displaystyle E \cdot ln(b) = ln(a)$

$\displaystyle e^{E \cdot ln(b)} = e^{ln(a)}$

The LHS can be re-arranged:

$\displaystyle e^{E \cdot ln(b)} = \left ( e^{ln(b)} \right ) ^E = b^E$

So

$\displaystyle b^E = a$

In the solution of your problem b = (1 - c)(1 - ee(S)) The ee(S) term is tangled up in the Eth power. There is no exact solution to this equation.

You can still do the graphs, though. Instead of the usual y = f(x) thing you can graph points implicitly. For instance if you want a plot of ee(S) (on the y axis) vs. E (on the x axis) for a particular value of c, pick a value for ee(S), say 3. Then find out what E is for that value. Then plot the point (E, 3). It'll take a lot of graphing single points, but it can be done.

-Dan

Re: Rearrange for X - cant seem to make any sense of this problem

Oh sorry, I clearly did that wrong. I definitely should not be doing math while sick.

Re: Rearrange for X - cant seem to make any sense of this problem

Thanks guys for your answers!

Then my math is not that horrible, as I ended up with the same conclusion as you, topsquarck. I'll see if I can plot the graphs anyways with the method you described.

In the article describing these formulas they say;

"Figure 1 B was computer generated by relating the variables c and ee(P)

in eq 5 to a function of x for values of 0 5 x 5 1; c = 1 - x/2 - xE/2; and

ee(P) = (x - xE)/(2 - x - x E ) . Implicit functions (eq 5 and 6) were solved

by parametric representations."

In here eq 5 and 6 is the same as the equation I posted in the first post.

Maybe Ill figure out how to do this with MATLAB or something!

Re: Rearrange for X - cant seem to make any sense of this problem

Quote:

Originally Posted by

**Educated** Oh sorry, I clearly did that wrong. I definitely should not be doing math while sick.

Pffl. The only reason I can spot the problem that easily is due to the number of times I've done the same thing!

-Dan

Re: Rearrange for X - cant seem to make any sense of this problem

Quote:

Originally Posted by

**vaffel** Thanks guys for your answers!

Then my math is not that horrible, as I ended up with the same conclusion as you, topsquarck. I'll see if I can plot the graphs anyways with the method you described.

In the article describing these formulas they say;

"Figure 1 B was computer generated by relating the variables c and ee(P)

in eq 5 to a function of x for values of 0 5 x 5 1; c = 1 - x/2 - xE/2; and

ee(P) = (x - xE)/(2 - x - x E ) . Implicit functions (eq 5 and 6) were solved

by parametric representations."

In here eq 5 and 6 is the same as the equation I posted in the first post.

Maybe Ill figure out how to do this with MATLAB or something!

If all you need to do is sketch a graph for given values of c I'd recommend this little graphics plotter. It's not fancy but it gets the job done.

-Dan