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2 ^ x + 3 ^ x = 50
find x

Is there any proper way to solve this other than trial and error?
I'm swung toward it being impossible. Give me your thoughts.

Are you looking for integer x, or any real x?
Ill assume its the latter since there is clearly no integer solution.
And areyou looking for a solution for this particular problem only or general a^{x}+b^{x}=c?

I think he wants a method other than numerical trial and error.

I tried having both sides be the power of e and it went nicely but I never got the correct answer which is between 3 and 4.

Are you looking for integer x, or any real x?

I can assure you there is no integer x which satisfies this equation:)

http://en.wikipedia.org/wiki/Fermat's_last_theorem

I doubt there is a nice answer, in a closed form.

As you said, what this question really boils down to is to solvee^{ax}+e^{bx}=c} algebraically.
Iwould say, unless a and b are exceptionally nice, this is not gonna happen, due to the non-algebraic nature of the exponential function.

I can assure you there is no integer x which satisfies this equation

Well its not quite Fermat since there is no x such that c^x=50. But its trivially impossible anyway since even+odd=odd.

Sorry scratch what I said about algebraic closure.

It does indeed look like a transcendental equation, in which case you can only solve it by approximation, or sometimes guess and check.

But if b was a multiple of a, for example b=na, theres a better chance of a nicer answer.
let u=e^{ax}. So we have u+u^{n}=c.
Which is a nice polynomial in u.
For example. If the OP had 2^{x}+4^{x}=50,
this could be reduced to u^{2}+u-50=0.
So u=\frac{-1+-\sqrt{ (201)}}{2}
Ignore the negative u.
So x=\frac{\log u}{\log2}, for what its worth. Which is probably not much since it doesnt help the OPs problem. :shrug:

Come to think of it, given 2^{x}+3^{x}=50,
write it as e^{ax}+e^{bx}=50.
Which is e^{ax}+e^{ax*b/a}=50.
Which can be written u+u^{\frac{b}{a}}=50.
Multiply everything by u^a.
u^{a+1}+u^{b}=50u^{a}.

This isnt a vast improvement but at least its a polynomial.

EDIT: Sorry scratch that. a and b are not integers so its not a polynomial. Ihavent reduced the problem.

The solution is ~ 3.353.
But I am not able to get this with an algebraic procedure.

cant u solve it with matrixes?

What's a matrix?

cant u solve it with matrixes?

Matrices.
Singular: matrix.
Plural: matrices.
:rolleyes:

cant u solve it with matrixes?

And no, I don't think you can.

x = 3.352823623213308 (from 5 iterations of Newton's method)

/*
* Newton Solver for 2^x + 3^x = 50
* f(x) = 2^x + 3^x - 50 = 0
* f'(x) = 2^x * ln 2 + 3^x * ln 3
* x_(i+1) = x_i - f(x_i)/f'(x_i)
*
*/
#include <stdio.h>
#include <math.h>

double fun (double x) {
double f;
f = pow(2.0,x) + pow(3.0,x) - 50;
return f;
}

double dfun (double x) {
double f;
f = pow(2.0,x) * log(2.0) + pow(3.0,x) * log(3.0);
return f;
}

int main () {
int i;
double x = 3.5,fx,dfx;
for (i=1; i<=10; i++) {
fx = fun(x);
dfx = dfun(x);
printf("i= %d x= %.15f f(x)= %g df(x)= %f\n",i,x,fx,dfx);
x = x - fx / dfx;
}
}

Many equations are "impossible", e.g., x + ln(x) = 1.

See Wikipedia:implicit function

x = 3.352823623213308 (from 5 iterations of Newton's method)

/*
* Newton Solver for 2^x + 3^x = 50
* f(x) = 2^x + 3^x - 50 = 0
* f'(x) = 2^x * ln 2 + 3^x * ln 3
* x_(i+1) = x_i - f(x_i)/f'(x_i)
*
*/
#include <stdio.h>
#include <math.h>

double fun (double x) {
double f;
f = pow(2.0,x) + pow(3.0,x) - 50;
return f;
}

double dfun (double x) {
double f;
f = pow(2.0,x) * log(2.0) + pow(3.0,x) * log(3.0);
return f;
}

int main () {
int i;
double x = 3.5,fx,dfx;
for (i=1; i<=10; i++) {
fx = fun(x);
dfx = dfun(x);
printf("i= %d x= %.15f f(x)= %g df(x)= %f\n",i,x,fx,dfx);
x = x - fx / dfx;
}
}

Thanks Kayyam!

Almost all equations in one unknown are solvable numerically using Newton or other successive approximaton methods.

To be very picky, such methods are really systematic trial & error.

BTW: Most polynomials of order 5 or greater cannot be solved algebraically & require Newton or some other numerical method.

It is rather curious that the equation ......

2^x = 3^x

is quite easy to solve in closed form by invoking
the inverse function. Take log of both sides and
it falls out almost immediately.

x = Log 3 / [Log 2 - Log 3]

It is rather curious that the equation 2^x = 3^x is quite easy to solve in closed form by invoking the inverse function. Take log of both sides ...
The only solution is obviously zero, but OK: x\log 2 = x\log 3, or x(\log 3 - \log 2) = 0. Since \log 2 \ne \log 3, the only solution is x=0.
... and it falls out almost immediately. x = Log 3 / [Log 2 - Log 3]
How did that fall out? BTW, this is not the problem posed in the OP.

Sorry, I made a typo.
Meant to write ...
2^x = 3^(x+1)

One of the x's should be x+1 for my solution to be correct.
This is a standard problem for my students when studying
Log and Exponential equations.

Again, the simple ommision of the constant of 50 in the original
problem makes it very easy to solve in closed form.