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Mathematics 15 Online
OpenStudy (anonymous):

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

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

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

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

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

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

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

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):

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

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 :)

OpenStudy (anonymous):

Hope I was helpful here

OpenStudy (anonymous):

U r always

OpenStudy (anonymous):

but

OpenStudy (anonymous):

but ... too slow

OpenStudy (anonymous):

noooooooooooooooo

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

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?

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

OpenStudy (anonymous):

this all in my studies

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