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Interesting equation

  1. Aug 27, 2005 #1
    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.
     
  2. jcsd
  3. Aug 27, 2005 #2
    u can use the property of logarithma

    x = log a
    base b

    b^x = a
     
  4. Aug 27, 2005 #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
     
  5. Aug 27, 2005 #4

    HallsofIvy

    User Avatar
    Science Advisor

    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.
     
  6. Aug 27, 2005 #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
     
  7. Aug 29, 2005 #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:
     
  8. Aug 29, 2005 #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: Aug 29, 2005
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