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Mathematics 13 Online
OpenStudy (anonymous):

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)

OpenStudy (anonymous):

So far I have hte following: log((x)*(x^2-49))/....

OpenStudy (anonymous):

Above all, I'm not sure how to simplify the log12

OpenStudy (anonymous):

Or rather, how to find the denom. in general. Any ideas?

OpenStudy (anonymous):

Is it just (12/(x+7)?

hartnn (hartnn):

you know \(\log a+\log b = \log ab \\ also,\log a-\log b = \log a/b \)

OpenStudy (anonymous):

Yes, I'm aware of all that.

OpenStudy (anonymous):

Factorise (x^2 - 49) (hint: part of this can then be cancelled out)

OpenStudy (anonymous):

If you're aware, then use it. Put it within one log and simplify.

OpenStudy (anonymous):

Where are you getting (x^2 - 49) from, by the way?

OpenStudy (anonymous):

log(x^2-49)

hartnn (hartnn):

so log(x) + log(x^2-49) - log12 - log(x+7) = log(x* (x^2-9) /[ 12 * (x+7) ] ) got this ?

OpenStudy (anonymous):

Oh...nevermind, wrong part of the problem..sorry...

OpenStudy (anonymous):

Yeah...that's what I entered (i.e. the site's saying it's wrong)

OpenStudy (anonymous):

Or no...I entered 12/(x+7) as the denom.

OpenStudy (anonymous):

Since in the original expression, you're subtracting (x+7) from log12

hartnn (hartnn):

that is incorrect, u got how ? and you can simplify x^2-49 = x^2-7^2 =??

hartnn (hartnn):

- log12 - log(x+7) = - ( log12 log(x+7)) = -(log [12*(x+7)])

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Okay...but in this one, there are two logs preceded by -, so does that mean the denominator is a fraction?

hartnn (hartnn):

no, \(\log a-\log b-\log c= \log (a/bc)\)

hartnn (hartnn):

\(\log a-\log b-\log c=\log a -(\log b+\log c)= \log (a/bc)\)

hartnn (hartnn):

got it ?

OpenStudy (anonymous):

Yep. Thank you.

OpenStudy (anonymous):

So the denom is (12x+84)?

hartnn (hartnn):

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

OpenStudy (anonymous):

So the num. becomes (x*x-7)?

hartnn (hartnn):

yeah and denominator = ?

OpenStudy (anonymous):

12?

hartnn (hartnn):

yes.

OpenStudy (anonymous):

Do I need two sets of parentheses around the denom?

hartnn (hartnn):

so u finally get \(\log [x(x-7)/12]\) got this ?

OpenStudy (anonymous):

ohh....okay

OpenStudy (anonymous):

Yes...thank you soo much!!

OpenStudy (anonymous):

Have a good day/night hartnn and I appreciate all your help!!

hartnn (hartnn):

you too :) welcome ^_^

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