How do you solve #lnx + ln(x+4) = 6#?

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Answer 1 · 2015-10-29T00:59:03

#-2+-sqrt(4+e^6)#

Start of by rewriting the left hand side using properties of logarithms

#ln(x(x+4))=6#

Distribute the #x#

#ln(x^2+4x)=6#

Rewrite both sides in terms of the base #e#

#e^(ln(x^2+4x))=e^6#

Rewrite left hand side using properties of logarithms

#x^2+4x=e^6#

Substract #e^6# form both sides

#x^2+4x-e^6=0#

#(-4+-sqrt(4^2-4(1)(e^6)))/(2(1))#

#(-4+-sqrt(16+4e^6))/2#

#(-4+-sqrt(4(4+e^6)))/2#

#(-4+-sqrt(4)sqrt(4+e^6))/2#

#(-4+-2sqrt(4+e^6))/2#

#-2+-sqrt(4+e^6)#