If $x_{n} > x_{n - 1} > . . . > x_{2} > x_{1} > 1$ , then the value of ${log}_{x_{1}} {log}_{x_{2}} {log}_{x_{3}} . . . {log}_{x_{n}} x_{n}^{(x_{n - 1})^{.^{.^{. x_{1}}}}}$ will be

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Solution

The correct option is A $1$
Given, ${log}_{x_{1}} {log}_{x_{2}} {log}_{x_{3}} . . . {log}_{x_{n}} x_{n}^{(x_{n - 1})^{.^{.^{. x_{1}}}}}$
$= {log}_{x_{1}} {log}_{x_{2}} {log}_{x_{3}} . . . {log}_{x_{n - 1}} {log}_{x_{n}} x_{n}^{(x_{n - 1})^{.^{.^{. x_{1}}}}}, [∵ log a^{b} = b log a]$
$= {log}_{x_{1}} {log}_{x_{2}} {log}_{x_{3}} . . . {log}_{x_{n - 1}} x_{n - 1}^{(x_{n - 2})^{.^{.^{. x_{1}}}}}$
Continuing this process, we get
$= {log}_{x_{1}} {log}_{x_{2}} {log}_{x_{3}} x_{3}^{(x_{2})^{^{. x_{1}}}}$
$= {log}_{x_{1}} {log}_{x_{2}} x_{2}^{x_{1}}$
$= {log}_{x_{1}} x_{1}$
$= 1$

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If xn>xn−1>...>x2>x1>1 , then the value of logx1logx2logx3...logxnx(xn−1)...x1n will be

The correct option is A 1Given, logx1logx2logx3...logxnx(xn−1)...x1n=logx1logx2logx3...logxn−1logxnx(xn−1)...x1n,[∵logab=bloga]=logx1logx2logx3...logxn−1x(xn−2)...x1n−1Continuing this process, we get=logx1logx2logx3x(x2).x13=logx1logx2xx12=logx1x1=1

If $x_{n} > x_{n - 1} > . . . > x_{2} > x_{1} > 1$ , then the value of ${log}_{x_{1}} {log}_{x_{2}} {log}_{x_{3}} . . . {log}_{x_{n}} x_{n}^{(x_{n - 1})^{.^{.^{. x_{1}}}}}$ will be