Solve $(x+a)^{1/x} = b$ , for $x$

where $a$ & $b$ are real constant.

Do not use Lambert W-function in solution.

Instead of using Lambert W-function, there are solution steps look like "completing the square method" of quadratic equation, but this solution needs steps of "completing the super square" to solve the equation by manipulating powers, but when dealing with powers note that this equation can be also written as:

$e^{2ni\pi}(x + a)^{1/x} = b$ when dealing with powers

where $e^{2ni\pi} = 1$

why I add $e^{2ni\pi}$ to the equation?

because if I want to power both sides by $x$ variable to eliminate the lift $1/x$ power and I did not assume that $n = 0$ then the equation will be like this:

$e^{x*2ni\pi}(x + a) = b^x$

and will be more complicated

so for this reason consider that $n=0$ in the solution to eliminate $e^{x*2ni\pi}$.

  • 3
    $\begingroup$ And the reason why what you ask is possible is... ? $\endgroup$ – user228113 Jun 30 '16 at 18:31
  • $\begingroup$ sorry Sassatelli I could not get your question!! $\endgroup$ – Nasser Dawood Jul 1 '16 at 13:34
  • $\begingroup$ What G. Sassatelli asked you is: why do you believe that it is possible to solve your equation without Lambert's function? $\endgroup$ – Alex M. Jul 1 '16 at 14:02
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    $\begingroup$ The "super square root" function used below seems to be a rather anecdotal reformulation of Lambert W-function. Thus, @NasserDawood, I would advise to meditate more indepth the comments above. $\endgroup$ – Did Jul 2 '16 at 13:00
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    $\begingroup$ You missed the point: there is nothing new in the so-called super square root function since it is completely equivalent to the Lambert W function. Hence every s-s-r solution can be translated into a Lambert-W one and vice versa. "New method": no. "New ideas": zero. (Unrelated: Please use @.) $\endgroup$ – Did Jul 2 '16 at 21:15


The method of this solution depends on "Completing the super square" of left side of the equation to make the base equal its power for getting super square root as following:

By starting with.

$x + a = b^x$

Let $y = x + a$

Substitute $y$ in the equation

$y = b^{y-a} \Rightarrow y = \frac {b^y}{b^a} \Rightarrow b^a y = b^y \Rightarrow \frac {1}{b^a y} = b^{-y}$

Now powering both sides to $\frac {1}{b^a y}$ for eliminating $y$ in the right side and making same power same base for the left side.

$(\frac {1}{b^a y})^{\frac {1}{b^a y}} = (b^{-y} )^{\frac {1}{b^a y}}$

$(\frac {1}{b^a y})^{\frac {1}{b^a y}} = b^{-y*\frac {1}{b^a y}} = b^{\frac {-1}{b^a} }$

Getting the super square root of both sides

$\sqrt {(\frac {1}{b^a y})^{\frac {1}{b^a y}}}_{s} = {\sqrt {b^{\frac {-1}{b^a} }}}_{s}$

Where $\sqrt {(...)}_{s}$ is super square root, and because the base of the left side equal its power then it is easy to get super square root of it as below.

$\frac {1}{b^a y} = {\sqrt {b^{\frac {-1}{b^a} }}}_{s}$

$y=\frac {1}{b^a {\sqrt {b^{\frac {-1}{b^a} }}}_{s}}$

$x + a =\frac {1}{b^a {\sqrt {b^{\frac {-1}{b^a} }}}_{s}}$

Finally the general formula is

$x = -a + \frac {1}{b^a {\sqrt {b^{(\frac {-1}{b^a}) }}}_{s}}$

Example to test the formula:

$(x + 2)^{1/x} = 2$


$x = -2 + \frac {1}{2^2 {\sqrt {2^{(\frac {-1}{2^2}) }}}_{s}} = -2 + \frac {1}{4 {\sqrt {2^{(\frac {-1}{4}) }}}_{s}} = -2 + \frac {1}{4 {\sqrt {\frac {1}{2^{(\frac {1}{4})}}}}_{s}}$

By calculating super square root values we obtain two real values.

${\sqrt {\frac {1}{2^{(\frac {1}{4})}}}}_{s} =\frac {1}{16} = 0.0625$ …… (because $0.0625^{0.0625}=0.8409=\frac {1}{2^{(\frac {1}{4})}}$ )


${\sqrt {\frac {1}{2^{(\frac {1}{4})}}}}_{s} = 0.8067$ …… (because $0.8067^{0.8067} = 0.8409 = \frac {1}{2^{(\frac {1}{4})}}$)

$x_{1}=-2 + \frac {1}{4*\frac {1}{16}} = -2 + 4 = 2$

$x_{2}=-2 + \frac {1}{4*0.8067}=-2 + 0.3099 = -1.6901$


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