Interesting equation

  • Thread starter Aphex_Twin
  • Start date
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.
 
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
 

HallsofIvy

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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? :yuck:
 
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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