anonymous
  • anonymous
log abc^0.5+log abc =?
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
options are: 5,9,2,1
xapproachesinfinity
  • xapproachesinfinity
let me see if this what you wrote before we begin \[\log abc^{\frac{1}{2}}+\log abc\]
xapproachesinfinity
  • xapproachesinfinity
is that right?

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

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