If `(log)_(10)2=a ,(log)_(10)3=bt h e n(log)_(0. 72)(9. 6)` in terms of `a` and `b` is equal to (a) `(2a+3b-1)/(5a+b-2)` (b) `(5a+b-1)/(3a+2b-2)` (c) `(3a+b-2)/(2a+3b-1)` (d) `(2a+5b-2)/(3a+b-1)`

Explanation

`log_0.72^9.6`

`log_10^9.6/log_10^0.72=(log_10^96-log_10^10)/(log_10^72-log_10^100)`

`=(log_10 (2^5*3)-1)/(log_10 (2^3*3^2)-2)`

`=(log_10 2^5+log_10 (3-1))/(log_10 2^3 +log_10 3^2 -2)`

`=(5a+b-1)/(3a+2b-2)`.