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Angelina.Paris@gmail.com1 science forum beginner
Joined: 17 Jul 2006
Posts: 2

Posted: Mon Jul 17, 2006 3:51 am Post subject:
Solving exponential inequality: a^x + b = c^x ???



Hi, sorry if this is trivial, but how can I solve equation in the
following form:
a^x + b = c^x
I'd appreciate any help.
Thanks 

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James Waldby science forum Guru Wannabe
Joined: 01 May 2005
Posts: 114

Posted: Mon Jul 17, 2006 4:29 am Post subject:
Re: Solving exponential inequality: a^x + b = c^x ???



Angelina.Paris wrote:
Quote: 
Hi, sorry if this is trivial, but how can I solve equation in the
following form:
a^x + b = c^x

Of a, b, c, and x, which ones are given, which unknown,
and which to be solved for? Are values integers,
reals, complex, or what?
jiw 

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moriman science forum beginner
Joined: 06 Apr 2006
Posts: 26

Posted: Mon Jul 17, 2006 5:18 am Post subject:
Re: Solving exponential inequality: a^x + b = c^x ???



<Angelina.Paris@gmail.com> wrote in message
news:1153108295.110898.223620@35g2000cwc.googlegroups.com...
Quote:  Hi, sorry if this is trivial, but how can I solve equation in the
following form:
a^x + b = c^x

Maybe you could ask James Harris to factor it, so it will be easier : =)
Quote:  I'd appreciate any help.
Thanks



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Angelina.Paris@gmail.com1 science forum beginner
Joined: 17 Jul 2006
Posts: 2

Posted: Mon Jul 17, 2006 6:07 am Post subject:
Re: Solving exponential inequality: a^x + b = c^x ???



James Waldby wrote:
Quote:  Angelina.Paris wrote:
Hi, sorry if this is trivial, but how can I solve equation in the
following form:
a^x + b = c^x
Of a, b, c, and x, which ones are given, which unknown,
and which to be solved for? Are values integers,
reals, complex, or what?
jiw

Sorry,
a, b, c are all real numbers
x is unknown that I need to solve this for. 

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Patrick Coilland science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197

Posted: Mon Jul 17, 2006 7:00 am Post subject:
Re: Solving exponential inequality: a^x + b = c^x ???



Angelina.Paris@gmail.com nous a récemment amicalement signifié :
Quote:  Hi, sorry if this is trivial, but how can I solve equation in the
following form:
a^x + b = c^x
I'd appreciate any help.
Hello 
In order a^x and c^x being meaningful, you need a > 0 and b > 0.
I suggest to consider a not equal to 1 (such a case is trivial).
Then you can write :
a^x + b = (a^x)^(ln(c)/ln(a))
So you have to solve in R+*
y + b = y^d
Except in some trivial cases (d=0, d=2, d=3, d=4, d=1, d=2, d=3,
d=1/2, d=1/3, d=1/4, d=1/2, d=3/2, d=2/3, and some others), I think
there is no general method to solve this.
You can then use a lot of approximations methods to approach a solution
(when it exists, because not all values of (b,d) lead to a solution).

Patrick 

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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Mon Jul 17, 2006 8:26 am Post subject:
Re: Solving exponential inequality: a^x + b = c^x ???



On Sun, 16 Jul 2006 Angelina.Paris@gmail.com wrote:
Quote:  James Waldby wrote:
Angelina.Paris wrote:
a^x + b = c^x
a, b, c are all real numbers
x is unknown that I need to solve this for.
You cannot solve that equation any more than you can solve 
x = a + sin x
Use numberial approximations or look up the Lambert(??) W function. 

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Robert B. Israel science forum Guru
Joined: 24 Mar 2005
Posts: 2151

Posted: Mon Jul 17, 2006 9:40 am Post subject:
Re: Solving exponential inequality: a^x + b = c^x ???



In article <Pine.BSI.4.58.0607170122510.9663@vista.hevanet.com>,
William Elliot <marsh@hevanet.remove.com> wrote:
Quote:  On Sun, 16 Jul 2006 Angelina.Paris@gmail.com wrote:
James Waldby wrote:
Angelina.Paris wrote:
a^x + b = c^x
a, b, c are all real numbers
x is unknown that I need to solve this for.
You cannot solve that equation any more than you can solve
x = a + sin x
Use numberial approximations or look up the Lambert(??) W function.

Numberial?
LambertW won't help here.
On the other hand, (after transforming to z + b = z^d as Patrick
suggested) you might look at the sci.math thread "Roots of Polynomials"
from JuneJuly 2004 to see how to get a solution as a series in
powers of b:
z = 1 + sum_{j=1}^infty pochhammer((j1)d/(1d),j1)/(j!(d1)) b^j
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 

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Ioannis science forum Guru Wannabe
Joined: 24 Mar 2005
Posts: 246

Posted: Mon Jul 17, 2006 5:28 pm Post subject:
Re: Solving exponential inequality: a^x + b = c^x ???



"Robert Israel" <israel@math.ubc.ca> wrote in message
news:e9fltl$loe$1@nntp.itservices.ubc.ca...
Quote: 
In article <Pine.BSI.4.58.0607170122510.9663@vista.hevanet.com>,
William Elliot <marsh@hevanet.remove.com> wrote:
On Sun, 16 Jul 2006 Angelina.Paris@gmail.com wrote:
James Waldby wrote:
Angelina.Paris wrote:
a^x + b = c^x
a, b, c are all real numbers
x is unknown that I need to solve this for.
You cannot solve that equation any more than you can solve
x = a + sin x
Use numberial approximations or look up the Lambert(??) W function.
Numberial?
LambertW won't help here.

Indeed, Lambert's W function is of no help here, but certain generalizations
of W, called HyperLambert functions, (or HW functions for short) can solve
it as follows:
a^x + b = c^x <=>
a^x  c^x = b <=>
x*(a^x  c^x)/x = b <=>
x*e^{log((a^x  c^x)/x)} = b <=>
x = HW(log((a^x  c^x)/x);b) <=>
x = HW(log(a^x  c^x)  log(x);b)
In general, the solution will be complex.
For details, see:
http://ioannis.virtualcomposer2000.com/math/HWPaper.html
and particularly the last section of:
http://ioannis.virtualcomposer2000.com/math/HWFunc.html
which deals with explicit solutions of Kepler's equation.
[snip]
Ioannis 

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