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

- anonymous

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

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- anonymous

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

- amistre64

emun can do it :)

- anonymous

yeah

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## More answers

- anonymous

give me a minute to finish it

- anonymous

fast plz

- anonymous

##### 1 Attachment

- amistre64

i mighta missed it lo

- anonymous

not sure. That is what i got. Correct me if im wrong.

- anonymous

wrong

- amistre64

1.27614 maybe lol

- anonymous

no

- anonymous

1 , -1

- anonymous

you have to isolate x now

- anonymous

why
U made the power 1/3 to 3

- anonymous

answer me

- anonymous

ahhh ok. Yeah definitively im wrong

- anonymous

gonna try to solve it.

- anonymous

U R welcome

- anonymous

Sorry i modified that, but i didnt get 1, -1

- anonymous

so what did U get

- anonymous

hah

- anonymous

answer me

- anonymous

something that you would make fun of. I have to have diner now. I'll answer you later. just be pacient

- anonymous

Here you have

##### 2 Attachments

- anonymous

wrong

- anonymous

look

- anonymous

Can you realize in what part I mess up?

- anonymous

log 20/10 = log 2

- anonymous

I didt do that anywhere

- anonymous

I dont need that process

- anonymous

sorry
at the end

- anonymous

\[(3X-1)(3X+1) = 8\]

- anonymous

it must be

- anonymous

You are right, but the mistake is at the begining. Gonna try to solve it again

- anonymous

\[9 x ^{2 } + 1 = 8\]

- anonymous

\[9 X ^{2} = 9\]

- anonymous

X = 1 , -1
s.s = {1}

- anonymous

-1 ref..

- anonymous

##### 1 Attachment

- anonymous

yeah you were right :)

- anonymous

Hope I was helpful here

- anonymous

U r always

- anonymous

but

- anonymous

but ... too slow

- anonymous

noooooooooooooooo

- anonymous

\[10^{\log20} *10^{-1}\]

- anonymous

or
\[10^{\log20} - 10^{-1}\]

- anonymous

the first one, because \[a ^{m}a ^{n}=a ^{m+n}\]

- anonymous

clear?

- anonymous

ok

- anonymous

can you give me a medal please :) I think I deserve it

- anonymous

but how

- myininaya

##### 1 Attachment

- anonymous

^ *cough*

- anonymous

forget something myininaya?

- myininaya

lol the 1
(x-1)(x+1)=0

- anonymous

no, you are saying the solution is x=+-1 , are you sure :P

- myininaya

well its not -1 because we can't have log of negative number

- myininaya

so its just x=1
i was hoping he would know to do that but you are right i should had said that

- anonymous

but I know

- anonymous

this all in my studies

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