Extracting X from Y: A, B, C, D Known

Thread starter shlomitk
Start date May 7, 2008

AI Thread Summary

To extract X from the equation Y = (A-D)/(1 + (X/C)^B + D), the rearrangement leads to the expression (A-D)/Y - 1 - D = (X/C)^B. This indicates that X can be isolated by manipulating the known values of A, B, C, and D, along with the variable Y. The discussion emphasizes the importance of correctly rearranging the equation to solve for X. Participants express varying levels of understanding regarding the algebra involved. Overall, the focus remains on finding a clear method to extract X using the provided variables.

May 7, 2008

shlomitk

how to extraxt the X

Y = (A-D)/(1 + (X/C)^B + D)

A,B,C,D, ARE KNOWN
Y is variable (OD value)

x - ?

Thanks

Physics news on Phys.org

May 7, 2008

Dragonfall

Hint: 1 + (X/C)^B + D = (A - D)/Y

May 7, 2008

BrendanH

Do you at least agree that (A-D)/y -1-D=(x/C)^B ?&&& seeing the last post, I realize that it looks as though my contribution tends to bupkes...

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

View full post »