If `log_a(ab)=x` then `log_b(ab)` is equals to

Explanation

`log_(a) (ab) = x or log_(a) a+ log_(a) b = x or log_(a) b= x - 1`
Now, ` log_(b)(ab) = log_(b) a+ log_(b) b`
` = 1/(log_(a) b) + 1 = 1/(x-1) + 1 x/(x-1)`