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How do you solve #log(x) + log(2x + 1) = 1#?

1 Answer

Dec 27, 2015

#color(white)xxx_1=2#
#color(white)xxx_2=-5/2#

Explanation:

#color(white)xxlogx+log(2x+1)=1#

#=>logx(2x+1)=1#
#=>2x^2+x=10#
#=>2x^2+x-10=0#

#color(white)xxDelta=81#
#=>x_(1,2)=(-1+-9)/4#

#=>x_1=2#
#=>x_2=-5/2#

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