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Say I have an equation like

$$2 z \ln[\frac{a v_{2}^{2}}{b_{2}}]+(v_{1}-2v_{2}) - z \ln[\frac{a v_{1}^{2}}{b_{1}}]\geq0$$

eq=2*z*Log[a*v2^2/b2]+(v1-2*v2)-z*Log[a*v1^2/b1]>=0

I'd like to write this in terms of b2/b1

Which would be something like

$$\frac{b_{2}}{b_{1}} \leq \frac{a v_{1}^{2}}{b_{2}a^{2}v_{2}^{4}}e^{v_{1}-2v{2}}$$

Note, that it hasn't been solved for $\frac{b_{2}}{b_{1}}$, there is still a $b_{2}$ on the right hand side.

Is it possible to ask Mathematica to simply for the ratio of some two variables? I have more complex expressions I'd like to simplify in this way.

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eq = 2*z*Log[a*v2^2/b2] + (v1 - 2*v2) - z*Log[a*v1^2/b1] >= 0;

sol = First@Solve[b21 == b2/b1, b1];

eq2 = eq /. sol

(*   v1 - 2 v2 - z Log[(a b21 v1^2)/b2] + 2 z Log[(a v2^2)/b2] >= 0   *)

sred = Simplify[Reduce[eq2, b21, Reals]] /. b21 -> b2/b1

TraditionalForm[
sred //. Or -> 
Composition[(Column[#, Right, Background -> {{White, LightGray}}, 
   Frame -> All] &), List]]

enter image description here

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