Taking log on both sides we get \[a^{x}ln(x) = x^{a}ln(a)\] .
The RHS is clearly a negative number since \(a<1\) hence \(ln(a) < 0\) and \(x^{a}\) is positive.
The LHS is an increasing function and it can also easily be checked that RHS is decreasing.So they have only one intersection point. and \(x=a\) is a trivial solution and hence the only one!! :)
–
Eddie The Head
·
2 years, 6 months ago

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TopNewestTaking log on both sides we get \[a^{x}ln(x) = x^{a}ln(a)\] . The RHS is clearly a negative number since \(a<1\) hence \(ln(a) < 0\) and \(x^{a}\) is positive. The LHS is an increasing function and it can also easily be checked that RHS is decreasing.So they have only one intersection point. and \(x=a\) is a trivial solution and hence the only one!! :) – Eddie The Head · 2 years, 6 months ago

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– Aaron Paul · 2 years ago

a<1 does not imply ln(a) is negetiveLog in to reply

– Sharky Kesa · 2 years ago

Well, \(a\) is greater than 0 and less than 1. Therefore, a is negative.Log in to reply