solve :- log(cubic root of 3X-1) + log(cubic root of 3x+1) =log 20-1

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solve :- log(cubic root of 3X-1) + log(cubic root of 3x+1) =log 20-1

Mathematics
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\[\log \sqrt[3]{3X-1} + \log \sqrt[3]{3X+1} =\log 20-1\]
emun can do it :)
yeah

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

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