How do you solve #log_516 - log_5 2t = log_5 2#?

1 Answer
Daniel L.
Jun 18, 2015

The answer is #t=4#
I assume that the question should be #log_5 16-log_5 2t=log_5 2#

Explanation:

First you have to simplify the left side and write the difference of logarythms as a logarythm of a division:

#log_5 (16/(2t))=log_5 2#

Now, when you have just logarythms on both sides and they both have the same base (5) you can get rid of logarythms and write:

#16/(2t)=2#

#8/t=2#

#8=2t#
#t=4#

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