## anonymous 3 years ago Simplify the expression attached and write it as a single logarithm.. ***still confused on this topic... pls help? :)

1. anonymous

2. anonymous

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

3. 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.

4. anonymous

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

5. tkhunny

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

6. 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 :(

7. tkhunny

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

8. anonymous

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

9. tkhunny

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

10. anonymous

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

11. tkhunny

Good, now the x-2...

12. anonymous

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

13. tkhunny

One more. You're on a roll!

14. anonymous

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

15. anonymous

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

16. tkhunny

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

17. anonymous

was what i got right?? or no?

18. anonymous

or is it log(x)^-2 ?

19. 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)

20. anonymous

sorry my computer crashed... one sec :/

21. anonymous

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

22. 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 ]

23. anonymous

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

24. jim_thompson5910

(x+4)^−3 is really 1/[ (x+4)^3 ]

25. jim_thompson5910

same idea applies to (x)^-2

26. anonymous

ok so i get 1/x^-2 ?

27. jim_thompson5910

yep

28. jim_thompson5910

so put that all together

29. anonymous

so i get answer C right?? :)

30. anonymous

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

31. anonymous

@jim_thompson5910 :)

32. jim_thompson5910

yes it is D

33. anonymous

kk great!!! thx :)