Question 199060
{{{ln(x-4)+ln(x+5)=1}}} Start with the given equation.



{{{ln((x-4)(x+5))=1}}}  Combine the logs using the identity {{{ln(A)+ln(B)=ln(A*B)}}}



{{{(x-4)(x+5)=e^1}}} Rewrite the equation using the property: {{{log(b,(x))=y}}} ====> {{{x=b^y}}} (note: the natural log has base "e")



{{{(x-4)(x+5)=e}}} Raise "e" to the first power to get "e"



{{{(x-4)(x+5)=2.718}}} Replace "e" with its approximate decimal value



{{{x^2+x-20=2.718}}} FOIL


 
{{{x^2+x-20-2.718=0}}} Subtract 2.718 from both sides.



{{{x^2+x-22.718=0}}} Combine like terms.



Now use either a calculator or the quadratic formula to get the approximate possible solutions:


{{{x=-5.293}}} or {{{x=4.293}}}



However, since you cannot take the log of a negative number, this excludes {{{x=-5.293}}}.



So the only approximate solution is {{{x=4.293}}}