As illustrated here I have the following MRI (Nuclear Magnetic Resonance Imaging) equation which I would like to rearrange for T1:
$$ S = k\times[H]\times{\frac{sin(\alpha)\times(1-\epsilon^{\frac{-TR}{T_1}})}{1-cos(\alpha)\times\epsilon^{\frac{-TR}{T_1}}}}\times\epsilon^{\frac{-TE}{T_2}} $$
I assume you'd start with:
$$ {\frac{S}{k\times[H]\times\epsilon^{\frac{-TE}{T_2}}}} = {\frac{sin(\alpha)\times(1-\epsilon^{\frac{-TR}{T_1}})}{1-cos(\alpha)\times\epsilon^{\frac{-TR}{T_1}}}} $$
But at this point I'm stuck. I think there are some trigonometric relationships you can use here to get the ${\frac{-TE}{T_2}}$ term by itself. Perhaps some Euler "trick"?
Edits / suggestions / migration / deletion welcome.
Assuming that you want to rearrange for $T_1$, considering$$S = k\,[H]\,{\frac{\sin(\alpha)\,(1-\epsilon^{-\frac{TR}{T_1}})}{1-\cos(\alpha)\,\epsilon^{-\frac{TR}{T_1}}}}\,\epsilon^{-\frac{TE}{T_2}}\tag 1$$ let us change notations $$s=\sin(\alpha)\qquad c=\cos(\alpha)\qquad x=\epsilon^{-\frac{TR}{T_1}}$$ which makes $$S= k\,[H]\,{\frac{s\,(1-x)}{1-c\,x}}\,\epsilon^{-\frac{TE}{T_2}}\tag 2$$ So $$\frac {S \epsilon^{\frac{TE}{T_2}} }{k[H]s}=\frac{1-x}{1-cx}$$ Using $$K=\frac {S \epsilon^{\frac{TE}{T_2}} }{k[H]s}\implies K=\frac{1-x}{1-cx}\implies x=\frac{K-1}{c K-1}$$
So $$T_1=-\text{TE}\,\frac{ \log (\epsilon )}{\log (x)}$$