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Solving exponential inequality: a^x + b = c^x ???
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Angelina.Paris@gmail.com1
science forum beginner


Joined: 17 Jul 2006
Posts: 2

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

Thanks
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James Waldby
science forum Guru Wannabe


Joined: 01 May 2005
Posts: 114

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

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

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

<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

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

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

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

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

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

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

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

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
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Ioannis
science forum Guru Wannabe


Joined: 24 Mar 2005
Posts: 246

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

"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 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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