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Given log(x^2-9)-log(x^2+7x+12) ----(*)
\n" ); document.write( "(x^2-9)=(x+3)(x-3)----(1)
\n" ); document.write( "(x^2+7x+12) =(x+4)(x+3)----(2)
\n" ); document.write( "Putting (1) and (2) in (*)
\n" ); document.write( "log(x^2-9)-log(x^2+7x+12) ----(*)
\n" ); document.write( "=log[(x+3)(x-3)]-log[(x+4)(x+3)]
\n" ); document.write( "=[log(x+3)+log(x-3)]-[log(x+4)+log(x+3)]
\n" ); document.write( "(using log(ab) = loga +logb all to the same base of course)
\n" ); document.write( "=log(x+3)+log(x-3)-log(x+4)-log(x+3)
\n" ); document.write( "=[log(x+3)-log(x+3)]+[log(x-3)-log(x+4)]
\n" ); document.write( "(using additive commutativity and associativity)
\n" ); document.write( "=0 +log[(x-3)/(x+4)](using loga-logb =log(a/b) all to the same base of course)
\n" ); document.write( "=log[(x-3)/(x+4)]
\n" ); document.write( "Answer:log[(x-3)/(x+4)]
\n" ); document.write( "Note:observe that
\n" ); document.write( "(x^2-9)is a quadratic in x of the form (a^2-b^2)
\n" ); document.write( "which is(a+b)(a-b)where a =x and b= 3]
\n" ); document.write( "Also observe that (x^2+7x+12) is a quadratic in x
\n" ); document.write( "which gives on factorisation(x+4)(x+3)
\n" ); document.write( "(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.
\n" ); document.write( "That is here 7x = (4x+3x) and (4x)X(3x) = 12x^2 = (x^2)X(12) \r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "