How do you solve #log_6x-log_6(x+6)=1#?

1 Answer
Euan S.
Jul 11, 2016

#x=-7.2#

Explanation:

Use rules of logs on left hand side:

#log_6((x)/(x+6)) = 1#

#x/(x+6) = 6#

#x = 6x +36 implies - 5x = 36#

#therefore x=-7.2#

If you sub that into your calculator to check you'll get an error - this is because we need a complex valued logarithm to deal with this! It isn't necessary to solve the question but you might be interested enough to look it up

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