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)
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