anonymous
  • anonymous
Help pleaseeeeeeeeeeee =*(
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1435701198421:dw|
anonymous
  • anonymous
\[\sqrt{\sqrt{\sqrt{3x}}}=\sqrt[6]{3x}\]\]
anonymous
  • anonymous
think that should be a 8 not 6

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anonymous
  • anonymous
so you have \[\sqrt[8]{3x}=\sqrt[4]{2x}\] raise each side to the power of 8
anonymous
  • anonymous
yes
anonymous
  • anonymous
yeah you are right it should be \(\sqrt[8]{2x}\) raise each side to the power of 8 to clear the radicals
anonymous
  • anonymous
\[3x=4x^2\] is what you will get then solve for \(x\)
anonymous
  • anonymous
hey how did u get that....?
anonymous
  • anonymous
raised both sides to the power of 8
anonymous
  • anonymous
wait what? why?
anonymous
  • anonymous
\[\sqrt[8]{3x}^8=3x\] \[\sqrt[4]{2x}^8=(2x)^2=4x^2\]
anonymous
  • anonymous
why? to get rid of the radicals
anonymous
  • anonymous
but i don't understand why when the base is different numbers
anonymous
  • anonymous
if you want to get rid of the racial on \(\sqrt[8]{3x}\) it should be pretty clear you have to raise it to the power of 8 right?
anonymous
  • anonymous
lol "radical"
anonymous
  • anonymous
no i don't understand that part is that like a rule or something?
anonymous
  • anonymous
when r u allowed to do that? when they're both divisible by the same number?
anonymous
  • anonymous
how else can you do it? you have the eighth root if you want to get it without the eighth root you have to raise it to the power of 8
anonymous
  • anonymous
what else can you do?
anonymous
  • anonymous
so you just come up with the 8 out of thin air? there isn't a rule or anything?
anonymous
  • anonymous
no , it was the eighth root if it was the fifteenth root, you would have to do something different
anonymous
  • anonymous
.... omg im so lost and nervous
anonymous
  • anonymous
I'm seriously not understanding that part
anonymous
  • anonymous
how are you going to get the \(3x\) outside of the radical ? you have \[\sqrt[8]{3x}\] what can you do to get rid of that radical?
anonymous
  • anonymous
if you square it, you would still have a radical wouldn't you ?
anonymous
  • anonymous
|dw:1435702119282:dw|
anonymous
  • anonymous
actually \(\left(3x\right)^{\frac{1}{8}}\)
anonymous
  • anonymous
okay so how come u get rid of that one but not the right side with the 2x? how come it gets neglected?
anonymous
  • anonymous
oh don't neglect it , it will feel left out if you raise the left hand side to the power or 8 you have to do the same thing to the right hand side
anonymous
  • anonymous
yes but the right still has a power of 2
anonymous
  • anonymous
\[\sqrt[4]{2x}^8=(2x)^2=4x^2\]
anonymous
  • anonymous
i thought we're trying to eliminate the power
anonymous
  • anonymous
you can't always get what you want you gotta use what you got
anonymous
  • anonymous
oops \[3x=4x^2\]
anonymous
  • anonymous
ok... i think i get it.... so what if the right side is not divisible by the number it's divided by on the right side?
anonymous
  • anonymous
on the left side no the rights *
anonymous
  • anonymous
that is an excellent question, best one yet
anonymous
  • anonymous
then i guess you would have to multiply by the least common multiple of the indices
anonymous
  • anonymous
oh ... i'm sorry to be so bothersome but can u please give me n example ? i just really wanna get this question and prepare for future problems like this
anonymous
  • anonymous
so if you had, for example \[\sqrt[4]{3x}=\sqrt[6]{2x}\] you would have to raise each side to the power of 12 since the least common multiple of 4 and 6 is 12
anonymous
  • anonymous
then both radicals would be gone and you would ge t \[(3x)^3=(2x)^2\]
anonymous
  • anonymous
ohhhhh hmmmmm
anonymous
  • anonymous
so the objective is to get rid of the fraction power right?
anonymous
  • anonymous
just so happens in this case raising to the power of 8 gets rid of the radical on both sides
anonymous
  • anonymous
yes
anonymous
  • anonymous
ahhhhhhhhhhhhhhhhhhhh i got it!!!!!!! thank you <333333333333333
anonymous
  • anonymous
yw
anonymous
  • anonymous
btw you still have to solve \[3x=4x^2\]
anonymous
  • anonymous
i got the answer i just needed to know how to do it
anonymous
  • anonymous
one answer is obviously 0 the other you get from solving \[4x^2-x=0\] etc
anonymous
  • anonymous
oops \[4x^2-3x=0\]
anonymous
  • anonymous
3/4 = .75
anonymous
  • anonymous
yes
anonymous
  • anonymous
Hey wait one more question.. if the problem was reversed it doesn't matter right? like the 3x was on the right side and the 2x was on the left as long as one of fractions is gone that's the objective no matter the left or right side right?
anonymous
  • anonymous
yeah that has nothing to do with it of course the answer would be different, but the method would be the same
anonymous
  • anonymous
okay gotcha thank you again <333

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