How do you solve #logx+log(x+3)=1#?

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Answer 1 · 2015-11-27T02:24:56

#x=2#

Know that: #{(loga+logb=logab),(logx=log_10x),(log_ab=c=>a^c=b):}#

#logx+log(x+3)=1#

#log(x^2+3x)=1#

#x^2+3x=10^1#

#x^2+3x-10=0#

#(x+5)(x-2)=0#

#color(red)cancel(x=-5)# or #x=2#

(For all #loga,a>0#.)