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Solving exponential equations in C++

Posted on 2009-05-07
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Last Modified: 2012-06-21
Is there a function or a library that can solve equations like: exp(constant*x) + exp(constant*x) = constant ?
If not , what is the best way to solve this equation in c++ ?

Thanks
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Question by:Chrysaor
8 Comments
 
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Expert Comment

by:tculler
ID: 24326232
Depends on what you mean by "exp"--what is this function supposed to do? Also, you can't assign a constant to anything but a constant, as far as I know. The entire expression must be a constant (unless x is a constant as well). Can you be a  bit more specific as to what exactly you're trying to accomplish?
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Expert Comment

by:jhshukla
ID: 24346957
He is trying to solve the equation "e^mx + e^nx = k" for x; m, n & k are known.
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Author Comment

by:Chrysaor
ID: 24350226
Excactly as jhshukla said.. Any ideas?
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Assisted Solution

by:MatrixDud
MatrixDud earned 332 total points
ID: 24496538
If all values are known, as you mentioned jhshukla was right about, then I don't see what you are trying to do (function is already solved).

The standard library has the exp function (math.h). You'll have to break the algebra down to its simplest form and solve for what ever variable you're looking for. Standard C++ libraries won't do that for you. A library like MathCAD might. Unless you need to break down a lot of different equations then you would be better to just do it yourself.
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Expert Comment

by:JIEXA
ID: 24533968

/* Newton's method example.
   http://en.wikipedia.org/wiki/Newton%27s_method */
#include <stdio.h>
#include <math.h>
 
/* the function */
#define f(x)	exp(2*x)+exp(3*x)-5
/* the derivative on x */
#define f_(x)	2*exp(2*x)+3*exp(3*x)
/* the first quess */
#define x0	-1
/* precision */
#define epsilon	1e-9
 
int main()
{
    double prev,curr,delta;
    prev=curr=x0;
    while(1)
    {
	curr = prev - (f(prev)/f_(prev));
	delta = fabs(curr-prev);
	printf(">> %.10f e=%.10f\n",curr,delta);
	if (delta < epsilon)
	  break;
	prev = curr;
    }
    printf("%.10f\n",curr);
    return 0;
}

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Assisted Solution

by:JIEXA
JIEXA earned 336 total points
ID: 24534003
Sorry, wrong usage of macros, here is new code.
/* Newton's method example.
   http://en.wikipedia.org/wiki/Newton%27s_method */
#include <stdio.h>
#include <math.h>
 
/* the function */
#define f(x)	exp(2*x)+exp(3*x)-5
/* the derivative on x */
#define f_(x)	2*exp(2*x)+3*exp(3*x)
/* the first guess */
#define x0	-1
/* precision */
#define epsilon	1e-9
 
int main()
{
    double prev,curr,delta;
    prev=curr=x0;
    while(1)
    {
	curr = prev - (f(prev))/(f_(prev));
	delta = fabs(curr-prev);
	/*printf(">> %.10f e=%.10f\n",curr,delta);*/
	if (delta < epsilon)
	  break;
	prev = curr;
    }
    printf("root=%.10f, f(root)=%.10f\n",curr,(f(curr)));
    return 0;
}

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Assisted Solution

by:evilrix
evilrix earned 332 total points
ID: 24534010
>> Sorry, wrong usage of macros, here is new code.
Just a style tip, in C++ prefer to use in-line functions, macros have so many reasons for not being a good idea in C++ it's not even funny :)
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Accepted Solution

by:
jhshukla earned 1000 total points
ID: 24542526
substitute e^x = A
A^m + A^n = k
depending on the values of m & n, the equation may be unsolvable or you may end up with non-real roots (for A).

x = ln(A)

wikipedia has a variety of formulas for solving cubic & quartic equations
http://en.wikipedia.org/wiki/Cubic_equation
http://en.wikipedia.org/wiki/Quartic_equation
there is no set formula for solving higher order equations. you can use newton's method as jiexa showed.
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