How do you condense $Log_b 2 + Log_bX- Log_by$ ?

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Answer 1 · 2016-05-27T13:52:10

Combine logs of the same base to get to $log_b((2X)/y)$

We can combine logs with the same base in this way:

$log_b2+log_bX-log_by$

$log_b((2X)/y)$

How do you condense Log_b 2 + Log_bX- Log_byLog_b 2 + Log_bX- Log_by?