# SOLUTION: Write log(x^2-9)-log(x^2+7x+12) as a single logarithim.

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 Algebra: Logarithm Solvers Lessons Answers archive Quiz In Depth

 Question 28454: Write log(x^2-9)-log(x^2+7x+12) as a single logarithim. Answer by sdmmadam@yahoo.com(530)   (Show Source): You can put this solution on YOUR website!Given log(x^2-9)-log(x^2+7x+12) ----(*) (x^2-9)=(x+3)(x-3)----(1) (x^2+7x+12) =(x+4)(x+3)----(2) Putting (1) and (2) in (*) log(x^2-9)-log(x^2+7x+12) ----(*) =log[(x+3)(x-3)]-log[(x+4)(x+3)] =[log(x+3)+log(x-3)]-[log(x+4)+log(x+3)] (using log(ab) = loga +logb all to the same base of course) =log(x+3)+log(x-3)-log(x+4)-log(x+3) =[log(x+3)-log(x+3)]+[log(x-3)-log(x+4)] (using additive commutativity and associativity) =0 +log[(x-3)/(x+4)](using loga-logb =log(a/b) all to the same base of course) =log[(x-3)/(x+4)] Answer:log[(x-3)/(x+4)] Note:observe that (x^2-9)is a quadratic in x of the form (a^2-b^2) which is(a+b)(a-b)where a =x and b= 3] Also observe that (x^2+7x+12) is a quadratic in x which gives on factorisation(x+4)(x+3) (x^2+7x+12) = (x^2+4x+3x+12) Writing the midterm as a sum of two terms in such a way that the product of the two terms is equal to the product of the square term and the constant term. That is here 7x = (4x+3x) and (4x)X(3x) = 12x^2 = (x^2)X(12)