the value of x when `log_(x) 343=3`,is

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

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))}]

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))}]

What is the value of x if log_(3)x+log_(9)x+log_(27)x+log_(81)x=(25)/(4)?

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

The following steps are involved in finding the value (7+x)^3 , when (7x)^3=343 . Arrange in sequential order. (A) (7+x)^3=(7+1)^3=8^3=512 (B) x^3=(343)/(7^3)=(7^3)/(7^3)=1 (C) rArrx=1 (D) (7x)^3=(343)/(7^3)=(7^3)/(7^3)=1