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If xn > x(n-1) > ..........> x3 > x1 > 1...

If `x_n > x_(n-1) > ..........> x_3 > x_1 > 1.` then the value of `log_(x1) [log_(x2) {log_(x3).........log_(x4) (x_n)^(x_(r=i))}]`

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`logx_1[logx_2{logx_3.........log_(x4)^(xr=1)}]`
`log_(x1)[log_(x2){log_(x3)....(xn)^(xr=1)log_(xn)^(xr=1)`
`log_(xn)[log_(x2)^(xr=1)]`
`log_(xr)^(x_1)`
`=1`
Option A is correct.
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