How do you solve $log_3 y – log_3 7 = log_3 2$ ?

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Answer 1 · 2015-06-25T15:08:04

Subtracting logs is dividing the arguments

$log_3 y-log_3 7=log_3 (y/7)->$
$log_3 (y/7)=log_3 2->$
$y/7=2->y=14$

This will work with all logs, as long as all suffixes are the same.

How do you solve log_3 y – log_3 7 = log_3 2log_3 y – log_3 7 = log_3 2?