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Basic of Logarithms
medium
If ${x_n} > {x_{n - 1}} > ... > {x_2} > {x_1} > 1$ then the value of ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}.....{\log _{{x_n}}}{x_n}^{x_{n - 1}^{{ {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} ^{{x_1}}}}}$ is equal to
A
$0$
B
$1$
C
$2$
D
None of these
Solution
(b) ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}………{\log _{{x_n}}}x_n^{x_{n – 1}^{{x^{{.^{{.^{{.^{{x_1}}}}}}}}}}}$
= ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}……..{\log _{{x_{n – 1}}}}x_{n – 1}^{x_{n – 2}^{{.^{{.^{{.^{{x_1}}}}}}}}}$
= ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}……{\log _{{x_{n – 2}}}}x_{n – 2}^{x_{n – 3}^{{.^{{.^{{.^{{x_1}}}}}}}}}$
= ${\log _{{x_1}}}{\log _{{x_2}}}x_2^{{x_1}} = {\log _{{x_1}}}{x_1} = 1$.
Standard 11
Mathematics
