Question 1008787
<pre>
{{{matrix(2,3,
"","","",

x^(1/log((x))),""="",k)}}}

x cannot be 1 since the denominator of the exponent
would be 0.

Take logs of both sides:

{{{matrix(2,3,
"","","",

log((x^(1/log((x))))),""="",log((k)))}}}

Use a rule of logs to write the exponent as a coefficient
of a log:

{{{matrix(2,3,
"","","",

(1/log((x)))log((x)),""="",log((k)))}}}

{{{matrix(2,3,
"","","",

(1/cross(log((x))))cross(log((x))),""="",log((k)))}}}

{{{matrix(2,3,
"","","",

1,""="",log((k)))}}}

{{{matrix(2,3,
"","","",

10,""="",k)}}}

Substitute for k in

{{{matrix(2,3,
"","","",

(1/log((x)))log((x)),""="",log((k)))}}}

{{{matrix(2,3,
"","","",

(1/log((x)))log((x)),""="",log((10)))}}}

{{{matrix(2,3,
"","","",

log((x))/log((x)),""="",1)}}}

This will be true for every allowable value of x

Thus the equation

{{{matrix(2,3,
"","","",

x^(1/log((x))),""="",k)}}}

has a solution for every positive value of x except 1

That is, for all x &#8712; {x|0 < x < 1 or x > 1}

 (0,1) U (1, &#8734;)

Edwin</pre>