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Answer 1 · 2017-12-05T01:10:35

#x=3/5 or 0.6#

#log_5 (5x+2)=1#

raise both sides to the 5

#5^(log_5(5x+2))=5^1#

#cancel(5^(log_5))(5x+2)=5^1#

#5x+2=5#

#5x=3#

#x=3/5=0.6#

Answer 2 · 2017-12-05T08:43:17

#x=3/5#

Using the definition of logarithms'

#log_ab=c=>a^c=b#

so

#log_5(5x+2)=1#

#=>5^1=5x+2#

solving for #x#

#5x=5-2=3#

#x=3/5#