Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Simplify the expression attached and write it as a single logarithm.. ***still confused on this topic... pls help? :)

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
answer choices A,B,C,D from top to bottom :)
Move all the exponents up: \((x+4)^{-3}\) \((x-7)^{2}\) \((x-2)^{5}\) \((x^{2})^{-1}\) Notice how I deliberately left everything outside as ADDITION. This allows us just to bring everything inside without concern for numerator or denominator. We can figure all that out later.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

im confused.. so how can i apply that with my expression?
Confused with what? \(-3\log(x+4) = \log\left[\left(x+4\right)^{-3}\right]\) Correct?
im not quite sure what we are working on now.. I'm like uber confused.. are we doing my problem? or an example? :/ sorry I'm a bit confused :(
Your problem statement begins \(-3\log(x+4)\), doesn't it? Or am I looking at some other picture?
yup iy does :) so i get log[(x+4)^−3] from that part?
Very good. Now the (x-7) part?
2log(x-7) = log(x-7)^2 ??
Good, now the x-2...
5log(x-2)=log(x-2)^5 ??
One more. You're on a roll!
lol thanks haha :P but i don't get this one... it looks different.. :/ -logx^2...
but heres my take on it.. log(x^2)^-1 ??
Take the negative (-1) inside, just like the other coefficients ==> exponents.
was what i got right?? or no?
or is it log(x)^-2 ?
There it is. They are now all connected by addition. Use this guy and bring them all together. log(a) + log(b) = log(a*b)
sorry my computer crashed... one sec :/
log[(x+4)^−3]+ log(x-2)^5+log(x-7)^2 + log(x)^-2=answer B right? :)
use the rule described above to go from log[(x+4)^−3]+ log(x-2)^5+log(x-7)^2 + log(x)^-2 to log[ (x+4)^−3*(x-2)^5*(x-7)^2*(x)^-2 ]
kk I'm seeing that :) but how can i simplify it even further?
(x+4)^−3 is really 1/[ (x+4)^3 ]
same idea applies to (x)^-2
ok so i get 1/x^-2 ?
yep
so put that all together
so i get answer C right?? :)
sorry i meant answer D :) is it answer D then? :)
yes it is D
kk great!!! thx :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question