Question 195711
{{{log(4,(x+3))+log(4,(x-3))=2}}} Start with the given equation.



{{{log(4,((x+3)(x-3)))=2}}} Combine the logs using the identity {{{log(b,(A))+log(b,(B))=log(b,(A*B))}}}



{{{(x+3)(x-3)=4^2}}} Rewrite the equation using the property: {{{log(b,(x))=y}}} ====> {{{x=b^y}}}



{{{(x+3)(x-3)=16}}} Square 4 to get 16



{{{x^2-9=16}}} FOIL



{{{x^2=16+9}}} Add 9 to both sides.



{{{x^2=25}}} Combine like terms.



{{{x=""+-sqrt(25)}}} Take the square root of both sides.



{{{x=sqrt(25)}}} or {{{x=-sqrt(25)}}} Break up the "plus/minus"



{{{x=5}}} or {{{x=-5}}} Take the square root of 25 to get 5




So the <i>possible</i> solutions are {{{x=5}}} or {{{x=-5}}}




However since you cannot take the log of a negative number, this rules out {{{x=-5}}} as a solution (since plugging this value in results in a negative argument)


================================================


Answer:


So the only solution is {{{x=5}}}