An algebra problem by حسن العطية

Algebra Level 4

If ${ x }^{ { x }^{ { x }^{ { . }^{ { . }^{ . } } } } }=n$

and $n>1$ is an odd integer,

then find $x$ .

{ x }=n

there is no such

x

{ x }=\sqrt [ \infty ]{ n }

{ x }=\sqrt [ n ]{ n }

1 solution

حسن العطية
Apr 22, 2016

${ x }^{ { x }^{ { x }^{ { . }^{ { . }^{ . } } } } }=n$

${ x }^{ n }=n$

$\boxed{{ x }=\sqrt [ n ]{ n }}$

Wrong! Except in the trivial case $n=1$ , there is no solution. See the discussion here

Otto Bretscher - 5 years, 1 month ago

you are right thank you for tell me

حسن العطية - 5 years, 1 month ago

Shall I change it to "no solution"?

Otto Bretscher - 5 years, 1 month ago

حسن العطية - 5 years, 1 month ago

Otto Bretscher - 5 years, 1 month ago