Math
posted by Halle .
Hi there
I need help with the restrictions on the variables of this question:
Simplify. State any restrictions on the variables.
log(x^2+7x+12)/log(x^29)
So my answer is: log(x+4/x3) which is correct.
Now for the restrictions, I have:
x<4 and x>3
However the back of the book says the answer is only x>3 and not x<4 for the restrictions of the variables. But when I sub in any number such as 5 into the ORIGINAL log equation, i get a defined and positive log of x value. So? is mine correct or theirs? Explain fully.

Sorry the question is:
log[(x^2+7x+12)/x^29)] 
first of all
log(x^2+7x+12)/log(x^29) is NOT equal to
log((x+4)/(x3)) like you say
If the back of your book has that answer, then they are definitely WRONG
log [(x^2+7x+12)/(x^29)]
would be equal to log((x+4)/(x3))
to find the restrictions you have to consider the numerator and denominator
remember we can only take logs of positive numbers
at the top:
log(x^2+7x+12) = log[(x+3)(x+4)]
x < 4 OR x > 3
at the bottom:
log(x^2  9) = log[(x3)(x+3)]
x < 3 or x > 3
but both of these conditions must be met, so
x < 4 OR x > 3 AND x < 3 or x > 3
which leaves you with x < 4 or x > 3
notice that between 4 and +3 we get a negative for either one or the other function
another way to understand the restriction would be:
graph both y = (x+3)(x+4) and
y = (x+3)(x3)
any x where either or both of those graphs dip below the xaxis would not be allowed.
so yes, they are wrong with their restrictions. 
ARGHHHH! saw your correction only after I posted.
So my answer would match your first posting, and it would clearly be a more interesting question. 
so now
log [(x+4)(x+3)/((x3)(x+3))]
= log(x+4) + log(x+3)  log(x+3)  log(x3)
now each of those terms must be defined.
so x>4 AND x>3 AND x>3
which is x > 3
notice that we canceled out log(x+3)
but for all x ≤ 3 that term would be undefined. So can we really do algebra with undefined numbers??
I know this is a bit tricky, since 5 (or any x < 4) would give a positive result in the original algebraic expression, but it would not work in the individuals terms.