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

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In summary, to solve an equation of the form a^x + x^b = c, where x is the unknown and a, b, and c are real constants, one can use the property of logarithms or a numerical method such as Newton's method. A potential solution involves using the Lambert's W-function, which is defined as the inverse function to f(x) = xex. The process involves simplifying the equation and using the W-function to solve for x. However, it is important to pay attention to signs and possible typos in the process.
  • #1
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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  • #2
u can use the property of logarithma

x = log a
base b

b^x = a
 
  • #3
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
 
  • #4
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.
 
  • #5
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
 
  • #6
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? :yuck:
 
  • #7
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:

What is the general method for solving equations of the form a^x + x^b = c?

The general method for solving equations of this form involves isolating the variable x on one side of the equation and using logarithms to solve for its value.

Can equations of this form have more than one solution?

Yes, equations of this form can have multiple solutions, depending on the values of a, b, and c. In some cases, there may be no real solutions.

What restrictions are there on the values of a, b, and c in this type of equation?

The values of a and b must be positive, and c must be a positive real number. Additionally, a and b cannot both be equal to 1.

Do I always need to use logarithms to solve equations of this form?

No, in some cases, it may be possible to solve these equations algebraically without using logarithms. However, using logarithms is a reliable method for solving these equations.

Are there any special cases I should be aware of when solving equations of this form?

Yes, if a = 1, then the equation simplifies to x + x^b = c, which can be solved using algebraic methods. Additionally, if b = 1, then the equation can be rewritten as a^x + x = c, which may be solved using numerical methods.

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