If ${x^{x\root 3 \of x }} = {(x\,.\,\root 3 \of x )^x},$ then $x =$
$64/27$
$-1$
$0$
None of these
The greatest number among $\root 3 \of 9 ,\root 4 \of {11} ,\root 6 \of {17} $ is
${{12} \over {3 + \sqrt 5 - 2\sqrt 2 }} = $
If ${a^{x - 1}} = bc,{b^{y - 1}} = ca,{c^{z - 1}} = ab,$then $\sum {(1/x) = } $
If ${a^x} = bc,{b^y} = ca,\,{c^z} = ab,$ then $xyz$=
${a^{m{{\log }_a}n}} = $