Solving Equations of a^x + x^b = c

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Discussion Overview

The discussion revolves around solving the equation of the form a^x + x^b = c, where x is the unknown and a, b, c are real constants. Participants explore various methods and approaches to tackle this equation, including analytical and numerical techniques.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests using logarithmic properties to express x in terms of a, b, and c.
  • Another participant proposes that the Lambert W-function might be applicable, along with numerical methods like Newton's method for finding solutions.
  • A participant attempts to manipulate the equation into a more solvable form but expresses uncertainty about reaching a dead end.
  • Further exploration leads to a more complex manipulation involving the Lambert W-function, but the participant questions if they made a sign error in their calculations.
  • Another participant acknowledges a miswriting of variables during their derivation, leading to confusion in their expressions.

Areas of Agreement / Disagreement

The discussion contains multiple competing views and approaches, with no consensus on a single method or solution. Participants express uncertainty and challenge each other's calculations without resolving the disagreements.

Contextual Notes

Participants' approaches depend on various assumptions about the values of a, b, and c, and there are unresolved mathematical steps in their derivations. The complexity of the equation leads to different interpretations and manipulations.

Aphex_Twin
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How does one proceed to solve an equation of the form:

a^x + x^b = c

Where x is the unknown (real or complex) and a, b, c real constants.

Or at least if you know a line of attack.
 
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u can use the property of logarithma

x = log a
base b

b^x = a
 
There must be something I'm missing, or unknowing of, that I reach a dead end:

a^x + x^b = c
x^x = c - x^b
x*ln(a) = ln(c - x^b)
x = ln(c-x^b)/(ln(a))
x = log_a (c-x^b)
??

where log_a is the base-a logarithm
 
Lambert's W-function? (Which is defined as the inverse function to f(x)= xex.)

Other than that, a numerical solution such as Newton's method.
 
I'm looking into it and it's very interesting. Please correct me if my calculations are wrong.

I start with a simpler equation, namely: x=2*ln(x)
e^x = x^2
1 = x^2/e^x
1 = x^2 * e^(-x)
1 = x * e^(-x/2)
-1/2 = -x/2 * e^(-x/2)

Therefore x = W(-1/2)


So going ahead with the main equation:

x = ln(c-x^b) * 1/(ln(a))
1 = ln(c-x^b) * (1/ln(a)) * 1/x
ln(a) = ln (c - x^b) * 1/x
a = (c - x^b) * e^(1/x)

I am stuck again
 
Eureka!


a^x+x^b=c
a^x = c - x^b
x*ln(a) = ln(c - x^b)
x = ln(c-x^b)/(ln(a))
x = ln(c-x^b) * (1/(ln(a)))
1 = ln(c-x^b) * (1/ln(a)) * 1/x
ln(a) = ln (c - x^b) * 1/x
ln(a) = ln(c-x^n)*e^(-ln(x))
(ln(a))^n = (ln(c-x^n))^n*e^(-ln(x^n))
(ln(a)^(n*ln(-1)) = (ln(c-x^n))^(n*ln(-1))*e^(-ln(-x^n))
(ln(a)^(n*ln(-1)*c) = ln(n*ln(-1)*c)*(c-x^n)*e^(-ln(c-x^n))
(ln(a))^(n*c*pi*i)=ln(n*c*pi*i)*(c-x^n)*e^(-ln(c-x^n))
ln(a) = ln(c-x^n)*e^((-ln(c-x^n)+1/(n*c*pi*i))
ln(a)=ln(c-x^n)*e^(1/(n*c*pi*i))*e^(-ln(c-x^n))
-ln(a)/(e^(1/(n*c*pi*i))) = -ln(c-x^n)*e^(-ln(c-x^n))
-ln(c-x^n) = W(-ln(a)/(e^(1/(n*c*pi*i))))
c-x^n = -e^(W(-ln(a)/(e^(1/(n*c*pi*i)))))
x^n = e^(W(-ln(a)/(e^(1/(n*c*pi*i))))) - c
x = (e^(W(-ln(a)/(e^(1/(n*c*pi*i))))) - c)^(1/n)

Did I lose a sign or something on the way?
 
whoops, I miswrote b as n somewhere on the mid way.

a^x+x^b=c
a^x = c - x^b
x*ln(a) = ln(c - x^b)
x = ln(c-x^b)/(ln(a))
x = ln(c-x^b) * (1/(ln(a)))
1 = ln(c-x^b) * (1/ln(a)) * 1/x
ln(a) = ln (c - x^b) * 1/x
ln(a) = ln(c-x^b)*e^(-ln(x))
(ln(a))^b = (ln(c-x^b))^b*e^(-ln(x^b))
(ln(a)^(b*ln(-1)) = (ln(c-x^b))^(b*ln(-1))*e^(-ln(-x^b))
(ln(a)^(b*ln(-1)*c) = ln(b*ln(-1)*c)*(c-x^b)*e^(-ln(c-x^b))
(ln(a))^(b*c*pi*i)=ln(b*c*pi*i)*(c-x^b)*e^(-ln(c-x^b))
ln(a) = ln(c-x^b)*e^((-ln(c-x^b)+1/(b*c*pi*i))
ln(a)=ln(c-x^b)*e^(1/(b*c*pi*i))*e^(-ln(c-x^b))
-ln(a)/(e^(1/(b*c*pi*i))) = -ln(c-x^b)*e^(-ln(c-x^b))
-ln(c-x^b) = W(-ln(a)/(e^(1/(b*c*pi*i))))
c-x^b = -e^(W(-ln(a)/(e^(1/(b*c*pi*i)))))
x^b = e^(W(-ln(a)/(e^(1/(b*c*pi*i))))) - c
x = (e^(W(-ln(a)/(e^(1/(b*c*pi*i))))) - c)^(1/b)
 
Last edited:

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