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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.
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Follow Ups
- I don't think this is the soln. you're looking for but it works... - Shane 13:46:28 07/28/00 (1)
- minor correction - Shane 13:51:09 07/28/00 (0)
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