How do you solve #log_5(2x+3) = log_5 3#?

1 Answer
Bdub
Mar 7, 2016

#x=0#

Explanation:

#log_5(2x+3)-log_5 3 =0# ->Put everything on one side
#log_5((2x+3)/3)=0#->Use property #log_b x-log_by=log_b(x/y)#
#5^0=1/3 (2x+3)#->Use definition #log_bx=y iff b^y=x#
#1=1/3 (2x+3)#
#3=2x+3#
#0=2x#
#x=0#

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