How do you solve #ln(x-6)+ln(x+3)=ln22#?

1 Answer
Aug 13, 2015

#color(red)( x=8)#

Explanation:

#ln(x-6)+ln(x+3)= ln22#

Recall that #lna+lnb=lnab#.

#ln(x-6)(x+3)=ln22#

Convert the logarithmic equation to an exponential equation.

#e^ln((x-6)(x+3)) = e^ln22#

Remember that #e^lnx =x#, so

#(x-6)(x+3)=22#

#x^2-3x-18=22#

#x^2-3x-40=0#

#(x-8)(x+5)=0#

#x-8=0# and #x+5=0#

#x=8# and #x=-5# are possible solutions.

Check:

#ln(x-6)+ln(x+3)= ln22#

If #x=8#,

#ln(8-6)+ln(8+3)=ln22#

#ln2+ln11=ln22#

#ln(2×11)=ln22#

#ln22=ln22#

#x=8# is a solution.

If #x=-5#,

#ln(-5-6)+ln(-5+3) =ln22#

#ln(-11)+ln(-2)=ln22#

But #ln(-11)# and #ln(-2)# are not defined.

#x=-5# is not a solution.