Question 130690
{{{log(4,(x+13))+log(4,(x+1))=3}}} Start with the given equation. 



{{{log(4,(x+13)(x+1))=3}}} Combine the logs using the identity {{{log(b,(A))+log(b,(B))=log(b,(A*B))}}}



{{{4^3=(x+13)(x+1)}}}  Rewrite the equation using the property: {{{log(b,(x))=y}}} ====> {{{b^y=x}}} 


{{{64=(x+13)(x+1)}}}  Evaluate {{{4^3}}} to get 64


{{{64=x^2+14x+13}}}  Foil the right side



{{{0=x^2+14x+13-64}}}  Subtract 64 from both sides. 


{{{0=x^2+14x-51}}}  Combine like terms



{{{(x+17)(x-3)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{x+17=0}}} or  {{{x-3=0}}} 


{{{x=-17}}} or  {{{x=3}}}    Now solve for x in each case



So our possible solutions are


 {{{x=-17}}} or  {{{x=3}}} 



However, if we plug in {{{x=-17}}}, we'll have negative numbers in the argument of the logs. Since you cannot take the log of a negative number, the value {{{x=-17}}} is an extraneous solution (ie it does not satisfy the original equation).




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Answer:


So our only solution is {{{x=3}}}