If {x^{xroot 3 of x }} = {(x,.,root 3 of x )^ | vedclass

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Question

(a) ${x^{x.{x^{1/3}}}} = {(x\,.\,{x^{1/3}})^x}$$ \Rightarrow $${x^{{x^{1 + {1 \over 3}}}}} = {\left( {{x^{1 + {1 \over 3}}}} \right)^x}$

==> ${x

Vedclass · Accepted Answer

$64/27$

Vedclass · Answer

(a) ${x^{x.{x^{1/3}}}} = {(x\,.\,{x^{1/3}})^x}$$ \Rightarrow $${x^{{x^{1 + {1 \over 3}}}}} = {\left( {{x^{1 + {1 \over 3}}}} ight)^x}$

==> ${x^{{x^{4/3}}}} = {\left( {{x^{4/3}}} ight)^x}$= ${x^{{x^{4/3}}}} = {x^{{4 \over 3}x}} \Rightarrow {x^{4/3}} = {4 \over 3}x$

$\Rightarrow {x^{\frac{4}{3} - 1}} = \frac{4}{3} \Rightarrow {x^{1/3}} = \frac{4}{3}$

 $	herefore x = {\left( {{4 \over 3}} ight)^3} = {{64} \over {27}}$

Also $x=1$ is an obvious solution .

If ${x^{x\root 3 \of x }} = {(x\,.\,\root 3 \of x )^x},$ then $x =$

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