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question attached inside!! :)

Mathematics
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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.

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part 1 of question :)
part 2 (last part!) of the question :) they are connected, but just one problem :)
A. -1 B. 27 C. 29 D. There isn't enough information to answer the question. idk :/

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Other answers:

all logs in this particular problem are base b
kk so that means? haha :P
log( (A^5C^2)/(D^6) ) = log(A^5C^2) - log(D^6) log( (A^5C^2)/(D^6) ) = log(A^5) + log(C^2) - log(D^6) What's next?
no i said that as a disclaimer to what I was about to write
oh okay haha :P so do i simplify that more? or is that the most simplified it can get?
or do i do substitution or sometihng?
you can pull the exponents down
using one of the rules given on that purplemath site
oh no!! i lost the link :(
can u pls resend it to me? :)
http://www.purplemath.com/modules/logrules.htm
okay so which one do i use?? I'm confused :(
you use the rule that log(x^y) = y*log(x)
so I'm trying to see what to put in here, but I'm stuck... idk :(
log( (A^5C^2)/(D^6) ) = log(A^5C^2) - log(D^6) log( (A^5C^2)/(D^6) ) = log(A^5) + log(C^2) - log(D^6) log( (A^5C^2)/(D^6) ) = 5*log(A) + 2*log(C) - 6*log(D) Then do the proper substitutions
ohhh i see what u did now! but how do i do proper substitutions??
log(A) = 5, so replace log(A) with 5 do the same with the other logs
so i get log(5)+log(2)-log(6) ??
close
you mixed up C and D though and you're missing your coefficients
okay then, idk then :( iheartfood=a confused person atm :/
look at the givens again
the givens?
what are the givens?? from the original question?
yeah, log(A) = ..., log(C) = ..., etc
okay so i get this? log(5)+log(7)-log(2) ??
ok, now add on the coefficients
what coefficients?? like the results when i enter it in the calc??
oh wait, idk why i said that was right, I'm misreading lol
log( (A^5C^2)/(D^6) ) = log(A^5C^2) - log(D^6) log( (A^5C^2)/(D^6) ) = log(A^5) + log(C^2) - log(D^6) log( (A^5C^2)/(D^6) ) = 5*log(A) + 2*log(C) - 6*log(D) log( (A^5C^2)/(D^6) ) = 5*5 + 2*7 - 6*2 log( (A^5C^2)/(D^6) ) = ???
log(A) = 5 NOT log(5)
wiat so what do i have now then?? I'm confused now lol
no sry, but i put up the right way to do it
so i have this so far?|dw:1353219981709:dw|??just making sure... idk if thats how I'm supposed to draw it?
hmm no
that looks off
can u pls draw it for me?? idk how to comprehend that well when its like in font/text format :/
\[\Large \log_{b}\left(\frac{A^5C^2}{D^6}\right) = \log_{b}\left(A^5C^2\right) - \log_{b}\left(D^6\right)\] \[\Large \log_{b}\left(\frac{A^5C^2}{D^6}\right) = \log_{b}\left(A^5\right) + \log_{b}\left(C^2\right) - \log_{b}\left(D^6\right)\] \[\Large \log_{b}\left(\frac{A^5C^2}{D^6}\right) = 5\log_{b}\left(A\right) + 2\log_{b}\left(C\right) - 6\log_{b}\left(D\right)\] \[\Large \log_{b}\left(\frac{A^5C^2}{D^6}\right) = 5(5) + 2(7) - 6(2)\] \[\Large \log_{b}\left(\frac{A^5C^2}{D^6}\right) = ???\]
27??
so my answer is B. 27 ???
yes
thank youuu :)

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