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RE: Inequality solving

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

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, 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])*

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}]

{{-(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] 
To: mathgroup at
Subject: [mg36977] [mg36966] Inequality solving


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.


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