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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
3 years, 3 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!! :) · 3 years, 3 months ago

a<1 does not imply ln(a) is negetive · 2 years, 9 months ago

Well, $$a$$ is greater than 0 and less than 1. Therefore, a is negative. · 2 years, 9 months ago