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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
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
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
 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
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.
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
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.
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 June-July 2004 to see how to get a solution as a series in
powers of b:

z = 1 + sum_{j=1}^infty pochhammer((j-1)d/(1-d),j-1)/(j!(d-1)) b^j

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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 , William Elliot 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 Hyper-Lambert 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]

 Quote: Robert Israel israel@math.ubc.ca --

Ioannis

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