How do you solve #log_3 6 + log_3x = log_3 12#?

1 Answer
EZ as pi
Sep 23, 2016

#x = 2#

Explanation:

If you are adding log terms, it means you can condense them into one log and multiply the numbers

#log_3 6 + log_3 x = log_3 12#

#log_3 (6 xx x) = log_3 12#

#log_3 6x = log_3 12" "larr if log A = log B, hArr A = B#

#6x = 12#

#x = 2#

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