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I would like to solve the equation:

A Exp[m t + s Sqrt[t] x - 1/2 s^2 t] >= K

with respect to x (which is in the exponent) using Mathematica. All parameters are positive. Of course, the answer is easy to get by hand. However, for some reason I could not figure out how to do it with Mathematica. What I tried gave me some error messages.

Also, I would like the solution to be expressed without all those conditionals. That is, I want the answer in which all the parameters satisfy whatever they need to satisfy so that the generic solution holds,

I would greatly appreciate help in figuring out how to do this.

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"gave me some error messages when I tried" — Please show us what you've tried. – R. M. Jan 18 '13 at 23:45
A kickstart Reduce[a Exp[m t + s Sqrt[t] x - 1/2 s^2 t] >= k, x, Reals] – Dr. belisarius Jan 18 '13 at 23:49
Note how @belisarius replaced K A with k a, this is because things starting with an upper-case letter are reserved for built-in functionality and it's good practice to avoid that or it will bite you in the ass – ssch Jan 18 '13 at 23:57
Tried it before, but get totally wrong solution. Have you tried it? Works OK if I replace Sqrt[t] with some variable say p, but still get huge mess of an expression (lots of extra conditions. – Branko Jan 18 '13 at 23:58
@ssch Not precisely true. Capital letters are used by a lot of third party packages, also, and there is nothing in the language that prevents this. It is a good recommendation, though, and if you decide not to follow it, you need to be careful. For the most part, I choose to ignore it, and capitalize my function names and lower case is used for my variable names. – rcollyer Jan 19 '13 at 3:02

If you replace k-> koa, a->1, Sqrt[t]->tt (since they're all positive parameters) then

Reduce[{ Exp[m tt^2 + s tt x - 1/2 s^2 tt^2] >= koa, m > 0, tt > 0, s > 0, koa > 0}, x, Reals]

(* koa > 0 && m > 0 && s > 0 && tt > 0 && 
   x >= (-2 m tt^2 + s^2 tt^2 + 2 Log[koa])/(2 s tt) *)
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