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Romania 1983 Question

Let \(0<a<1\). Solve

\(x^{a^x} = a^{x^a}\)

for positive numbers \(x\)

Note by Sharky Kesa
2 years, 9 months ago

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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, 9 months ago

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@Eddie The Head a<1 does not imply ln(a) is negetive Aaron Paul · 2 years, 3 months ago

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@Aaron Paul Well, \(a\) is greater than 0 and less than 1. Therefore, a is negative. Sharky Kesa · 2 years, 3 months ago

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