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Find the value of $x$ : $x^{x^{3}}=3$

I tied with "log" but I couldn't. any help?

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    $\begingroup$ It doesn't look like it has a nice solution, even with the W function. bit.ly/1JmHBGw I'm not sure what Zev has done but he probably assumed you meant $(x^x)^3$ instead of $x^(x^3)$; which did you mean? $\endgroup$
    – Jam
    Jun 26, 2015 at 0:48
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    $\begingroup$ @EulCan $\sqrt[3]{3}$ is in fact a solution to the latter form. $\endgroup$ Jun 26, 2015 at 1:29

4 Answers 4

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Hint: $\large (\sqrt[3]{3})^3=3{}{}$.

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  • $\begingroup$ brother, I hate hints, I tried but I couldn't. $\endgroup$
    – SAM
    Jun 26, 2015 at 0:43
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    $\begingroup$ Look again. @zev has all but given you the answer. $\endgroup$
    – Simon S
    Jun 26, 2015 at 0:44
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A nice first step toward the solution is to increase the symmetry of the LHS: if you cube $x^{x^3}$, you get $x^{3x^3}$, which is $(x^3)^{x^3}$. Symmetry! Since cubing the RHS gives $3^3$, it's easy to then spot the solution $x^3=3$.

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  • $\begingroup$ Very good approach! $\endgroup$
    – Integral
    Jun 26, 2015 at 1:34
  • $\begingroup$ well done. is the solution unique? $\endgroup$ Jun 26, 2015 at 2:53
  • $\begingroup$ @MichaelChirico : For that aspect, consult KfSsOc's solution. $\endgroup$
    – user21467
    Jun 26, 2015 at 4:17
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consider $$x^{x^b}=a :x>0$$

take log for both side: $$x^b\log x=\log a$$

let $e^t=x$

$$e^{bt}t=\log a$$ $$e^{at}bt=b\log a$$ so

$$bt=W(b\log a)$$

$$t=\frac{W(b\log a)}{b}$$

$$x=e^{\frac{W(b\log a)}{b}}$$

where $W(R)$ is Lambert W function

for $a=b=3$

$x \approx 1.44225 $

another try: $${\color{Red} c}=x^c \Rightarrow x=\sqrt[c]{c}$$

$$\Rightarrow x^{{\color{Red} {x^c}}}=c$$

$$ \Rightarrow x=\sqrt[c]{c}$$

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    $\begingroup$ ...or more simply $x = a^{1/a}$. $\endgroup$
    – Winther
    Jun 26, 2015 at 0:53
  • $\begingroup$ @Winther i agree with you so i will edit it $\endgroup$
    – mnsh
    Jun 26, 2015 at 0:58
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Assume the domain is $(0,\infty)$. You show the equation $x^{x^3} = 3$ has a unique real solution $x = \sqrt[3]{3}$. Look at the function: $f(x) = x^3\ln x , 0 < x < \infty$. If $0 < x < 1 \Rightarrow x^3\ln x < 0$, so $x^3\ln x < \ln 3$, since $\ln 3 > 0$. Thus: $e^{x^3\ln x} < e^{\ln 3}\Rightarrow x^{x^3} < 3$. And for $ 1\leq x <\infty$, the function $f(x) = x^3\ln x$ has $f'(x) = 3x^2\ln x+ x^2 = x^2(3\ln x + 1) > 0$. This means the equation $x^{x^3} = 3$ can only have atmost $1$ solution on $[1,\infty)$. Observe that $x = \sqrt[3]{3}$ is a solution, and so it is the only solution.

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