#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.
