Mathematics
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OpenStudy (anonymous):
solve :-
log(cubic root of 3X-1) + log(cubic root of 3x+1) =log 20-1
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OpenStudy (anonymous):
\[\log \sqrt[3]{3X-1} + \log \sqrt[3]{3X+1} =\log 20-1\]
OpenStudy (amistre64):
emun can do it :)
OpenStudy (anonymous):
yeah
OpenStudy (anonymous):
give me a minute to finish it
OpenStudy (anonymous):
fast plz
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OpenStudy (anonymous):
OpenStudy (amistre64):
i mighta missed it lo
OpenStudy (anonymous):
not sure. That is what i got. Correct me if im wrong.
OpenStudy (anonymous):
wrong
OpenStudy (amistre64):
1.27614 maybe lol
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OpenStudy (anonymous):
no
OpenStudy (anonymous):
1 , -1
OpenStudy (anonymous):
you have to isolate x now
OpenStudy (anonymous):
why
U made the power 1/3 to 3
OpenStudy (anonymous):
answer me
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OpenStudy (anonymous):
ahhh ok. Yeah definitively im wrong
OpenStudy (anonymous):
gonna try to solve it.
OpenStudy (anonymous):
U R welcome
OpenStudy (anonymous):
Sorry i modified that, but i didnt get 1, -1
OpenStudy (anonymous):
so what did U get
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OpenStudy (anonymous):
hah
OpenStudy (anonymous):
answer me
OpenStudy (anonymous):
something that you would make fun of. I have to have diner now. I'll answer you later. just be pacient
OpenStudy (anonymous):
Here you have
OpenStudy (anonymous):
wrong
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OpenStudy (anonymous):
look
OpenStudy (anonymous):
Can you realize in what part I mess up?
OpenStudy (anonymous):
log 20/10 = log 2
OpenStudy (anonymous):
I didt do that anywhere
OpenStudy (anonymous):
I dont need that process
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OpenStudy (anonymous):
sorry
at the end
OpenStudy (anonymous):
\[(3X-1)(3X+1) = 8\]
OpenStudy (anonymous):
it must be
OpenStudy (anonymous):
You are right, but the mistake is at the begining. Gonna try to solve it again
OpenStudy (anonymous):
\[9 x ^{2 } + 1 = 8\]
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OpenStudy (anonymous):
\[9 X ^{2} = 9\]
OpenStudy (anonymous):
X = 1 , -1
s.s = {1}
OpenStudy (anonymous):
-1 ref..
OpenStudy (anonymous):
OpenStudy (anonymous):
yeah you were right :)
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OpenStudy (anonymous):
Hope I was helpful here
OpenStudy (anonymous):
U r always
OpenStudy (anonymous):
but
OpenStudy (anonymous):
but ... too slow
OpenStudy (anonymous):
noooooooooooooooo
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OpenStudy (anonymous):
\[10^{\log20} *10^{-1}\]
OpenStudy (anonymous):
or
\[10^{\log20} - 10^{-1}\]
OpenStudy (anonymous):
the first one, because \[a ^{m}a ^{n}=a ^{m+n}\]
OpenStudy (anonymous):
clear?
OpenStudy (anonymous):
ok
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OpenStudy (anonymous):
can you give me a medal please :) I think I deserve it
OpenStudy (anonymous):
but how
myininaya (myininaya):
OpenStudy (anonymous):
^ *cough*
OpenStudy (anonymous):
forget something myininaya?
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myininaya (myininaya):
lol the 1
(x-1)(x+1)=0
OpenStudy (anonymous):
no, you are saying the solution is x=+-1 , are you sure :P
myininaya (myininaya):
well its not -1 because we can't have log of negative number
myininaya (myininaya):
so its just x=1
i was hoping he would know to do that but you are right i should had said that
OpenStudy (anonymous):
but I know
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OpenStudy (anonymous):
this all in my studies