If {x_n} {x_{n - 1}} ... {x_2} {x_1} | vedclass

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(B) Let the expression be $E = \log_{x_1} \log_{x_2} \log_{x_3} \dots \log_{x_n} (x_n^{x_{n-1}^{\dots^{x_1}}})$.Using the property $\log_a (a^b) = b$,

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(B) Let the expression be $E = \log_{x_1} \log_{x_2} \log_{x_3} \dots \log_{x_n} (x_n^{x_{n-1}^{\dots^{x_1}}})$.Using the property $\log_a (a^b) = b$,we evaluate from the innermost logarithm:$\log_{x_n} (x_n^{x_{n-1}^{\dots^{x_1}}}) = x_{n-1}^{x_{n-2}^{\dots^{x_1}}}$.Substituting this back into the expression:$E = \log_{x_1} \log_{x_2} \dots \log_{x_{n-1}} (x_{n-1}^{x_{n-2}^{\dots^{x_1}}}) = \log_{x_1} \log_{x_2} \dots \log_{x_{n-1}} (x_{n-1}^{x_{n-2}^{\dots^{x_1}}}) = \log_{x_1} \log_{x_2} \dots \log_{x_{n-2}} (x_{n-2}^{x_{n-3}^{\dots^{x_1}}})$.Continuing this process iteratively:$E = \log_{x_1} \log_{x_2} (x_2^{x_1}) = \log_{x_1} (x_1) = 1$.

If ${x_n} > {x_{n - 1}} > ... > {x_2} > {x_1} > 1$,then the value of ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}.....{\log _{{x_n}}}({x_n}^{{x_{n - 1}}^{{.^{{.^{{.^{{x_1}}}}}}}}})$ is equal to

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