If log0.3(x-1)0.09(x-1), then x lies in the interval:a.(2,∞)b. (1,2)c. (-2,-1)d. none of these(JEE 1983)Answer (a)log0.3(x-1)0.09(x-1),=> log (x-1)/log 0.3 < 09 =""> log (x-1)/log 0.3 < 3 =""> log (x-1)>1/2 log (x-1) (as log 0.3<0)> long (x-1)>0=> x-1>1=> x>2thus x Є (2,∞)