How do you solve #Log_6 (x+3) = 1- log_6 (x-2)#?

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Answer 1 · 2018-06-26T10:03:56

#x=3#

#log_6(x+3)=1+log_6(x-2)#

#log_6(x+3)=log_6 6+log_6(x-2)#

#log_6(x+3)=log_6 6(x-2)#

#x+3=6(x-2)#

#x+3=6x-12#

#5x=15#

#x=3#

Remember:
#log_a b+log_a c=log_a bc#
#log_a a=1#