If `0 lt a lt 1,` the value of `log_10 a` is negative. This is justified by

If 0 lt x lt 1 , the first negative term in the expansion of (1+x)^(27//5) is

If 0 lt x lt 1 , the first negative term in the expansion of (1+x)^(27//5) is

If a,b, and c are in G.P., with 1 lt a lt b lt x and n gt 1 is an integer then find the sequence formed by log_a n,log_b n,log_c n ……….

If 0 lt a lt 1 , then the solution set of the inequation (1+(log_(a)x)^(2))/(1+(log_(a)x)) gt1 , is

If 0 lt a lt 1 , then the solution set of the inequation (1+(log_(a)x)^(2))/(1+(log_(a)x)) gt1 , is

The maximum value of |x log x| for 0 lt x le 1 is

Statement - 1 : If x gt 1 then log_(10)x lt log_(3)x lt log_(e )x lt log_(2)x . Statement - 2 : If 0 lt x lt 1 , then log_(x)a gt log_(x)b implies a lt b .

Statement - 1 : If x gt 1 then log_(10)x lt log_(3)x lt log_(e )x lt log_(2)x . Statement - 2 : If 0 lt x lt 1 , then log_(x)a gt log_(x)b implies a lt b .