Logarithms - 6

If log_0.3(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)

log_0.3(x-1)0.09(x-1),

=> log (x-1)/log 0.3 < log (x-1)/log 0.09

=> log (x-1)/log 0.3 < log (x-1)/2log 0.3
=> log (x-1)>1/2 log (x-1) (as log 0.3<0)
=> long (x-1)>0
=> x-1>1
=> x>2
thus x Є (2,∞)