This comes from physics environment. Would anyone be able to help me alternate this equation in order to find the $x$?
I need to have this equation in the form of $x = \text{...something}$. $$ A = \frac{B}{C} - \Bigg( \frac{B}{C} - \frac{B}{D} \Bigg)e^{- \frac{BF}{X}} $$ Please note: $e$ is an Euler's number here, and this is $e$ to the power of minus $\frac{BF}{X}$ (looks little bit confusing).
My attempt:
$- \frac{A - \frac{B}{C}}{\frac{B}{C} - \frac{B}{D}} = e^{-\frac{BF}{X}}$
$\frac{D(AC-B)}{B(C-D)} = e^{-\frac{BF}{X}}$
$\ln{\Bigg(\frac{D(AC-B)}{B(C-D)}\Bigg)} = -\frac{BF}{X}$
$1 = -\frac{BF}{X\ln{\Bigg(\frac{D(AC-B)}{B(C-D)}\Bigg)}}$
$X = -\frac{BF}{\ln{\Bigg(\frac{D(AC-B)}{B(C-D)}\Bigg)}}$
However I'm not sure if it's correct.
Help's appreciated, thanks.
Consider the problem of
$$A=G-He^{-\frac{K}{X}}$$
We have
$$He^{-\frac{K}{X}}=G-A$$
$$e^{-\frac{K}{X}}=\frac{G-A}{H}$$
Hence
$$-\frac{K}{X}=\log\left(\frac{G-A}{H} \right)$$
Hopefully you can continue from here.
Edit:
$$\frac{K}{X}=\log\left( \frac{H}{G-A}\right)$$
Hence
$$X=\frac{K}{\log\left( \frac{H}{G-A}\right)}$$
Now, $K=BF, G=\frac{B}{C}, H=\frac{B}{C}-\frac{B}{D}$
Hence,
$$X=\frac{BF}{\log\left( \frac{\frac{B}{C}-\frac{B}{D}}{\frac{B}{C}-A}\right)}=\frac{BF}{\log\left( \frac{\frac{BD-BC}{CD}}{\frac{B-AC}{C}}\right)}=\frac{BF}{\log\left( \frac{B(D-C)}{D(B-AC)}\right)}$$