`log_a(1-1/2)+log_a(1-1/3)+......log_a(1-1/n)`

log2+log(1+(1)/(2))+log(1+(1)/(3))+.........+log(1+(1)/(n-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))}]

(1)/(log_(2)x)+(1)/(log_(3)x)+......(1)/(log_(43)x)=(1)/(log_(43)!x)

If xy^(2) = 4 and log_(3) (log_(2) x) + log_(1//3) (log_(1//2) y)=1 , then x equals

given k=log_(3)2 find the value of (1)/(log_(3)4)+(1)/(log_(3)4)+(1)/(log_(3)3)+......+(1)/(log_(3)15)

Find log_(10)1((1)/(2))+log_(10)1((1)/(3))+log_(10)1((1)/(4))+ upto 198 terms

Solve: log_(1/3) (x-1) > log_(1/3) 4