$log_{a} (ab)$ = x, then $log_{b} (ab)$ is :
1/x
x/ (x+1)
x/(1-x)
x/(x-1)
Explanation:
$log_{a} (ab)$ = x
=> log b/ log a = x
=> (log a + log b)/ log a = x
1+ (log b/ log a) = x
=> log b/ log a = x-1
log a/ log b = 1/ (x-1)
=> 1+ (log a/ log b) = 1 + 1/ (x-1)
(log b/ log b) + (log a/ log b) = x/ (x-1) => (log b + log a)/ log b = x/ (x-1)
=>log (ab)/ log b = x/(x-1) => $log_{b} (ab)$ = x/(x-1)