Solving for ${x}^{{x}^{3}}=5$

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I am trying to solve $${x}^{{x}^{3}}=5$$

I took logarithms on both sides, and got $${x}^{3}\cdot \ln \left( x \right) -\ln \left( 5 \right) =0$$

Then I took the help of Maple, when I solve for x the first equation, I get $$x={{\rm e}^{{\frac {{\rm W} \left(3\,\ln \left( 5 \right) \right)}{3} }}}$$

$W$ here is the Lambert W function.

and when I solve for $x$ after taking the logartihms (2nd equation) I get

$$ \left[ \begin {array}{c} x={{\rm e}^{{\it RootOf} \left( \left( { {\rm e}^{{\it \_Z}}} \right) ^{3}\cdot {\it \_Z}-\ln \left( 5 \right) \right) }}\end {array} \right] $$

Can anyone explain, why there is difference? My 2nd question is how do we come here to using Lambert function?