How do you solve #ln x + ln (x+6) = 1/2 ln 9#?

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Answer 1 · 2015-06-20T01:25:04

#x=-3+\sqrt{12}#
Use rules for adding logarithms and rule for bringing the coefficient inside the logarithm as an exponent.

#ln(A)+ln(B)=ln(AB)#

so

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

#Cln(D)=ln(D^C)#

so

#1/2ln(9)=ln(9^{1/2})=ln(3)#

Making these substitutions,

#ln(x)+ln(x+6)=1/2ln(9)#

becomes

#ln(x^2+6x)=ln(3)#

which requires,

#x^2+6x=3#

#\implies x^2+6x-3=0#

#x={-6\pm\sqrt{6^2-4(1)(-3)}}/{2(1)}#

#x=-3\pm\sqrt{12}#

#ln(x)# requires #x>0# so we choose the plus sign

#x=-3+sqrt{12}#