log abc^0.5+log abc =?

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log abc^0.5+log abc =?

Linear Algebra
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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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options are: 5,9,2,1
let me see if this what you wrote before we begin \[\log abc^{\frac{1}{2}}+\log abc\]
is that right?

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Other answers:

is it (abc)^0.5 or just ab(c^0.5) huge difference
so which is it?
hello!! you there?
yes the first one is ryt
will be 1/2log abc +log abc
so it is (abc)^1/2 yes
yes
okay then \(\log (abc)^{frac{1}{2}}+\log abc= \frac{1}{2} \log abc+\log abc\) \(=\frac{3}{2}\log(abc)\)
that first line where frac12 supposed to be all to 1/2
so we get \(=\log(abc)^{\frac{3}{2}}\)
yea , next step
that is all what is it you want to achieve?
a different way of doing that is \(\log (abc)^{\frac{1}{2}}+\log abc=\log \frac{(abc)^{\frac{1}{2}}}{abc}\)
but the options of this ques given are a) 5 b) 9 c) 2 d) 1 which onw wud be the ans?
i meant \(=log[(abc)^0.5 (abc)]\) not division!!
\(=\log[(abc)^{0.5} (abc)]\)
if those are your options then you missing something in the question
alright
can you post the picture of your question
that must be the wrong ques. i pick this ques from mformaths.
bdw thankew for your help. :)
that question cannot be any of those number no more simplification is required
yep
all you can do is something like this \(=3/2[\log a+\log b+ \log c]\) which pretty much does not help for anything new
ok i have another question cud u help me in that?
post is in a different post though
okz

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