`log(x)-log(2x-3)=1`, Then `x`

If 9^("log"3("log"_(2) x)) = "log"_(2)x - ("log"_(2)x)^(2) + 1, then x =

If 9^("log"_3("log"_(2) x)) = "log"_(2)x - ("log"_(2)x)^(2) + 1, then x =

2log x-log(x+1)-log(x-1)=

If f(x) {:(=(log (1-2x)-log(1-3x))/x", " x != 0 ),(=a", " x = 0 ):} is continuous at x = 0 , then : a =

Differentiate log(3x+2)-x^(2)log(2x-1) with respect to x:

If f(x)=tan^(-1)[(log((e )/(x^(2))))/(log (ex^(2)))]+tan^(-1)[(3+2 log x)/(1-6 log x)] then the value of f''(x) is

2log x-log(x+1)-log(x-1) is equals to

2log x-log(x+1)-log(x-1) is equals to

If log2,log(2^(x)-1),log(2^(x)+3) are in A.P. write the value of x.

log(x-1)+log(x-2)lt log(x+2)