Films/DVD Asylum

Posted by Shane (i) on July 28, 2000 at 13:46:28:

In Reply to: nope, he's from tsa-gi :) posted by gnat on July 27, 2000 at 23:51:29:

a^x=x^a
x*ln(a)=a*ln(x)
x/a=ln(x-a)
(e^x)/(e^a)=e^(ln(x-a))
(e^x)/(e^a)=x-a
(e^x)*(e^(-a))=x-a
e^(-a*x)=x-a
now take da/dx for both sides
(-a)*e^(-a*x)=1
e^(-a*x)=-a^-1
ln(e^(-a*x))=ln(-a^-1)
-a*x=-ln(-a)
a*x=ln(-a)
x=ln(-a)/a
since a>=0, ln(-a)=infinity
therefore x=infinity.
checking our work
a^x=x^a
a^infinity=infinity^a for all a>1
yes, x=infinity is other solution.