The value of `x, log_(1/2)x >= log_(1/3)x` is

The value of x satisfying log_(3)4 -2 log_(3)sqrt(3x +1) =1 - log_(3)(5x -2)

If x_n > x_(n-1) > ..........> x_3 > x_1 > 1. then the value of log_(x_1) [log_(x _2) {log_(x_3).........log_(x_n) (x_n)^(x_(r=i))}]

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))}]

If x_n > x_(n-1) > ..........> x_3 > x_1 > 1. then the value of log_(x_1) [log_(x _2) {log_(x_3).........log_(x_n) (x_n)^(x_(r=i))}]

If x_n > x_(n-1) > ..........> x_3 > x_1 > 1. then the value of log_(x_1) [log_(x _2) {log_(x_3).........log_(x_n) (x_n)^(x_(r=i))}]

Find the value of x,log_(4)(x+3)-log_(4)(x-1)=log_(4)8-2

If x = log_(3) 4 " and " y = log_(5) 3 , find the value of log_(3) 10 " and " log_(3) (1 . 2) in terms of x and y .

Find the value of x such that log_(0.3)(x-1) = log_(0.09)(x-1) .