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
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