RE: Inequality solving

*To*: mathgroup at smc.vnet.net*Subject*: [mg36977] RE: [mg36966] Inequality solving*From*: "DrBob" <drbob at bigfoot.com>*Date*: Thu, 3 Oct 2002 05:33:31 -0400 (EDT)*Reply-to*: <drbob at bigfoot.com>*Sender*: owner-wri-mathgroup at wolfram.com

Here's a start. If you want the thing to be greater than zero, set it equal to z (which we'll assume is greater than zero) and solve for Log[x]: Log[x, a] + Log[a*x, a] == z /. Log[a_*b_] -> Log[a] + Log[b] {f, g} = Log[x] /. Simplify[Solve[%, Log[x]]] Log[a]/Log[x] + Log[a]/(Log[a] + Log[x]) == z {-(((-2 + z + Sqrt[4 + z^2])*Log[a])/(2*z)), ((2 - z + Sqrt[4 + z^2])* Log[a])/(2*z)} Neither solution appears to be extraneous. Now the task is to find the range of these functions over positive z. Take a look at their derivatives: D[f, z] // Simplify D[g, z] // Simplify ((-1 + 2/Sqrt[4 + z^2])* Log[a])/z^2 ((-1 - 2/Sqrt[4 + z^2])*Log[a])/z^2 A little study shows that f' and g' have their signs opposite to Log[a]. Both functions are monotone. The following limits: Outer[Limit[#1, z -> #2] &, {f, g}, {0, Infinity}] Exp@% {{-(Log[a]/2), -Log[a]}, {Infinity*Log[a], 0}} {{1/Sqrt[a], 1/a}, {Indeterminate, 1}} show that f varies from -Log[a]/2 to -Log[a] and g varies from 0 to Infinity if Log[a]>0 and 0 to -Infinity if Log[a]<0. Exponentiation gives ranges for x: -1/Sqrt[a] to 1/a for f, for instance. But what does all this mean? When a>1, either x>1 or 1/a < x < 1/Sqrt[a]. When a<1, either 0<x<1 or 1/Sqrt[a] < x < 1/a. That is... if I haven't screwed something up. Bobby Treat -----Original Message----- From: CeZaR [mailto:pascal at go.ro] To: mathgroup at smc.vnet.net Subject: [mg36977] [mg36966] Inequality solving Hi, Can I solve this inequality with Mathematica? Log[x,a]+Log[a x,a]>0 I've tried all know, but I get get any good result. Thanks! CeZaR

**Follow-Ups**:**Peculiar output from DSolve***From:*Selwyn Hollis <selwynh@earthlink.net>