ln(x-6)+ln(x+3)= ln22ln(x−6)+ln(x+3)=ln22
Recall that lna+lnb=lnablna+lnb=lnab.
∴ ln(x-6)(x+3)=ln22ln(x−6)(x+3)=ln22
Convert the logarithmic equation to an exponential equation.
e^ln((x-6)(x+3)) = e^ln22eln((x−6)(x+3))=eln22
Remember that e^lnx =xelnx=x, so
(x-6)(x+3)=22(x−6)(x+3)=22
x^2-3x-18=22x2−3x−18=22
x^2-3x-40=0x2−3x−40=0
(x-8)(x+5)=0(x−8)(x+5)=0
x-8=0x−8=0 and x+5=0x+5=0
x=8x=8 and x=-5x=−5 are possible solutions.
Check:
ln(x-6)+ln(x+3)= ln22ln(x−6)+ln(x+3)=ln22
If x=8x=8,
ln(8-6)+ln(8+3)=ln22ln(8−6)+ln(8+3)=ln22
ln2+ln11=ln22ln2+ln11=ln22
ln(2×11)=ln22ln(2×11)=ln22
ln22=ln22ln22=ln22
x=8x=8 is a solution.
If x=-5x=−5,
ln(-5-6)+ln(-5+3) =ln22ln(−5−6)+ln(−5+3)=ln22
ln(-11)+ln(-2)=ln22ln(−11)+ln(−2)=ln22
But ln(-11)ln(−11) and ln(-2)ln(−2) are not defined.
∴ x=-5x=−5 is not a solution.