[x/(x-1)] log_{a} (ab) = x, then log

For Competitive Exams

$log_{a} (ab)$ = 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)

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