How do you solve #log_4 x =2-log_4 (x+6)#?

1 Answer
Bdub
Mar 7, 2016

#log_4x+log_4(x+6)=2->log_4(x*(x+6))=2->(log_4(x^2+6x))=2->4^2=x^2+6x->0=x^2+6x-16#
#(x+8)(x-2)=0->x=-8 and x=2# Ans: #x=2#

Explanation:

First, combine all the logs on one side then use definition to change from the sum of the logs to the log of a product. Then use the definition to change to exponential form and then solve for x. Note that we cannot take a log of a negative number so -8 is not a solution.

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