Suppose we have an expression like $ \ln x^{x^{x}} $. We can write it as $ \ln x^{x^{x}} = x^x\ln x $. But this can also be written as: $$ \ln x^{x^{x}} = \ln t^x; \ where \ t = x^x, \\ \implies \ln t^x = x\ln t = x \ln x^x = x.x \ln x = x^2 \ln x $$ I know I am wrong somewhere in my logic but where?
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You took the order of exponentiation wrongly. For example, $3^{3^3}$ should be computed as $3^{27}$, not $(3^3)^3= 27^3 = 3^9$. So $$\mathrm{ln} x^{x^x} = x^x \mathrm{ln}x.$$ – oleout Jun 28 '21 at 10:38
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Exponentiation is not associative: $$ x^{(x^x)} \not= (x^x)^x = x^{x\cdot x} = x^{x^2}. $$
This is why after taking logarithms you get different values.
Kolja
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Your error strarts from here:
$$x^{x^x}=x^{(x^x)}≠\left(x^x\right)^x=x^{x^2}=x^{(x^2)}.$$
lone student
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When you write $\ln x^{x^x} = x^x \ln x$, you read $x^{x^x}$ as $x^{(x^x)}$. When you get $x^2 \ln x$, you have read $x^{x^x}$ as $(x^x)^x$. The first one is correct.
Magdiragdag
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