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

1 Answer

Mar 11, 2016

#=>x=-5 and x=2#

Explanation:

#log(x+3)+logx =1#
#=>log((x+3)xxx) =log10#
#=>x^2+3x=10#
#=>x^2+3x-10=0#
#=>x^2+5x-2x-10=0#
#=>x(x+5)-2(x+5)=0#
#=>(x+5)(x-2)=0#
#=>x=-5 and x=2#

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