We have #xoy=x^(xlog_e y),forall x,yin[1,oo)#.Find #x# for #x o x o x=125#?

1 Answer
Jun 23, 2017

#x = e^root(4)(3 log 5)#

Explanation:

Considering that for #x > 0 rArr x = e^(log x)#

and defining #x@y = e^(logx logy)#

we have

#x@x@x = e^(Log(e^(Log(e^(Log^2x)) Logx)) Logx) = ((e^(Log^2x))^Logx)^Logx#

then

#((e^(Log^2x))^Logx)^Logx=5^3#

now applying #log # to both sides

#logx log(e^(Log^2x))^Logx=log^2x log(e^(Log^2x))=log^4x = 3 log 5#

then

#log x= root(4)(3 log 5)# and

#x = e^root(4)(3 log 5)#