Use properties of logarithms to condense the logarithmic expression. Write the expression as a single logarithm whose coefficient is 1: log(x) + log(x^2-49) - log12 - log(x+7)
So far I have hte following: log((x)*(x^2-49))/....
Above all, I'm not sure how to simplify the log12
Or rather, how to find the denom. in general. Any ideas?
Is it just (12/(x+7)?
you know \(\log a+\log b = \log ab \\ also,\log a-\log b = \log a/b \)
Yes, I'm aware of all that.
Factorise (x^2 - 49) (hint: part of this can then be cancelled out)
If you're aware, then use it. Put it within one log and simplify.
Where are you getting (x^2 - 49) from, by the way?
log(x^2-49)
so log(x) + log(x^2-49) - log12 - log(x+7) = log(x* (x^2-9) /[ 12 * (x+7) ] ) got this ?
Oh...nevermind, wrong part of the problem..sorry...
Yeah...that's what I entered (i.e. the site's saying it's wrong)
Or no...I entered 12/(x+7) as the denom.
Since in the original expression, you're subtracting (x+7) from log12
that is incorrect, u got how ? and you can simplify x^2-49 = x^2-7^2 =??
- log12 - log(x+7) = - ( log12 log(x+7)) = -(log [12*(x+7)])
Basically in a log equation, whatever has a + sign goes on the top (numerator) and whatever has a - goes on the bottom (denominator) As long as they are of the same base and coefficient
Okay...but in this one, there are two logs preceded by -, so does that mean the denominator is a fraction?
no, \(\log a-\log b-\log c= \log (a/bc)\)
\(\log a-\log b-\log c=\log a -(\log b+\log c)= \log (a/bc)\)
got it ?
Yep. Thank you.
So the denom is (12x+84)?
keep it as 12(x+7) because in numerator u get x^2-49 = (x+7)(x-7) so that u can cancel x+7
So the num. becomes (x*x-7)?
yeah and denominator = ?
12?
yes.
Do I need two sets of parentheses around the denom?
so u finally get \(\log [x(x-7)/12]\) got this ?
ohh....okay
Yes...thank you soo much!!
Have a good day/night hartnn and I appreciate all your help!!
you too :) welcome ^_^
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