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I am trying to solve the following inequality for $v$: $b>e^{\frac{-v}{\alpha}}\cdot \alpha + v$. (Where I know: $b>0$, $v \in [0, 1]$, and $\alpha >0$).

However the following doesn't work: Reduce[b>E^(-v/a)*a+v, v]. I get the error message "This system cannot be solved with the methods available to Reduce". Simplify, and Solve doesn't work either.

Could you think of an alternative? Or is it unsolvable?

Thank you.

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There is unlikely to be a "nice" closed-form solution for this equation, since it involves both transcendental and algebraic expressions for $v$. However, it can be solved in terms of special functions with a bit of work.

We can make the substitution $v = \alpha x + b$ to rewrite the equation (after some algebra) as $$ x e^x < - e^{-b/\alpha }. $$ Now, $x e^x$ is a monotonically increasing function so long as $x > -1$:

Reduce[D[x Exp[x], x] > 0, x]
(* x > -1 *)

Given the provided constraints on $\alpha$, $v$, and $b$, we will always have $x \geq 0$. This means that $x$ must be less than the solution to the equation $x e^x = - e^{-b/\alpha}$, and Mathematica can solve this in terms of a special function:

Solve[x Exp[x] == - Exp[-b/a], x]
(* {{x -> ProductLog[-E^(-(b/a))]}} *)

Thus, the solution to the inequality will be $x < W(-e^{-b/\alpha})$, where $w$ is the Lambert $W$-function (aka the product-log function.) In terms of $v$, this then means that $$ v < b + \alpha W(-e^{-b/\alpha}). $$

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    $\begingroup$ You may have dropped a sign. Starting with ineq1 = b > E^(-v/a)*a+v /. v -> a*x+b // ExpandAll; ineq2 = SubtractSides[ineq1, b+a*x]; ineq3 = MultiplySides[ineq2, -E^x/a, Assumptions -> a > 0 && Element[x, Reals]] evaluates to -E^(-(b/a)) > E^x*x $\endgroup$
    – Bob Hanlon
    Jul 21 '21 at 17:04
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    $\begingroup$ @BobHanlon: You're 100% correct. Thanks for the catch. $\endgroup$ Jul 21 '21 at 17:17

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