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\[\log \sqrt[3]{3X-1} + \log \sqrt[3]{3X+1} =\log 20-1\]

emun can do it :)

yeah

give me a minute to finish it

fast plz

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

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..

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

^ *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