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

- anonymous

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- schrodinger

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- anonymous

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- anonymous

answer choices A,B,C,D from top to bottom :)

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- tkhunny

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.

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## More answers

- anonymous

im confused.. so how can i apply that with my expression?

- tkhunny

Confused with what? \(-3\log(x+4) = \log\left[\left(x+4\right)^{-3}\right]\) Correct?

- anonymous

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 :(

- tkhunny

Your problem statement begins \(-3\log(x+4)\), doesn't it? Or am I looking at some other picture?

- anonymous

yup iy does :) so i get log[(x+4)^−3] from that part?

- tkhunny

Very good. Now the (x-7) part?

- anonymous

2log(x-7) = log(x-7)^2 ??

- tkhunny

Good, now the x-2...

- anonymous

5log(x-2)=log(x-2)^5 ??

- tkhunny

One more. You're on a roll!

- anonymous

lol thanks haha :P but i don't get this one... it looks different.. :/
-logx^2...

- anonymous

but heres my take on it..
log(x^2)^-1 ??

- tkhunny

Take the negative (-1) inside, just like the other coefficients ==> exponents.

- anonymous

was what i got right?? or no?

- anonymous

or is it log(x)^-2 ?

- tkhunny

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)

- anonymous

sorry my computer crashed... one sec :/

- anonymous

log[(x+4)^−3]+ log(x-2)^5+log(x-7)^2 + log(x)^-2=answer B right? :)

- jim_thompson5910

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 ]

- anonymous

kk I'm seeing that :) but how can i simplify it even further?

- jim_thompson5910

(x+4)^−3 is really 1/[ (x+4)^3 ]

- jim_thompson5910

same idea applies to (x)^-2

- anonymous

ok so i get 1/x^-2 ?

- jim_thompson5910

yep

- jim_thompson5910

so put that all together

- anonymous

so i get answer C right?? :)

- anonymous

sorry i meant answer D :) is it answer D then? :)

- anonymous

@jim_thompson5910 :)

- jim_thompson5910

yes it is D

- anonymous

kk great!!! thx :)

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