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Please help! I'm having trouble with logs. Click attached :)

Mathematics
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1 Attachment
do you know the laws of logs?
I'm faimiliar with them, yes

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Other answers:

*familiar
ok so one of them says that when you have log(a/b), it is the same as log(a)-log(b) ... you can apply that hear :)
are we okay?
Yes, I understand. Thanks!
you're welcome :)
what did you get though?
Didn't you post this same problem the other day @IloveCharlie
\(\text{Latex is working}\)
/(/large hi /) ... if this doesn't appear correctly, then i probably forgot how to write latex, if not, yay! else, :(
ok so it messed up.... what did i do wrong?
:/ I'm having trouble...
\(found \ what \ i \ did \ wrong \ \text{:P} \) anyway what are you having trouble with?
Is it log(3.8)-log(5)?
@Hero Hmm... no. If so, that'd be weird. I don't think my Precal/Trig online course would repeat problems. Maybe it was something similar?
u said what it equal to
actually, no. the question has a quotient/division thingy, which obviously means there's two sides to it. one is \(\large \sqrt{(3.2)(4.7)} \) and the other is \(\large 5 \) according to the law, \(\large log(\frac{a}{b})=log(a) - log(b) \) in this case, a= \(\large \sqrt{(3.2)(4.7)} \) and b = \(\large 5 \) hence: \(\large log(\frac{\sqrt{(3.2)(4.7)}}{5})=log(\sqrt{(3.2)(4.7)}) - log(5) \) =\(\large log(\sqrt{(15.04)} \ )-log(5) \)
are we good to go?
Ohhhh o.k. Now I get it. Thanks so much sasogeek
no problem :) you're welcome x

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