how do I solve this? Watch

SidTheSloth1
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#1
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I already got the answer, but it involved some guessing, what would be the best way to solve this (qb)
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aoxa
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Divide by four, write as surds then work backwards to get n on its own.


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SidTheSloth1
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Could you write out the steps one by one?, and how would you be able to write it as a surd?
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Kvothe the Arcane
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(Original post by SidTheSloth1)
Could you write out the steps one by one?, and how would you be able to write it as a surd?
Write 4 and 8 in terms of powers of 2. Take it from there. Recall the rules of indices.
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Matureb
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Start by changing the 8 and 4 into powers of 2.
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Kvothe the Arcane
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(Original post by aoxa)
Divide by four, write as surds then work backwards to get n on its own.


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I don't honestly see this as useful advice.
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SidTheSloth1
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So I would write 2^2n^3/2 = 2^3/8 ? I would understand better if someone wrote out each step
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davros
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(Original post by SidTheSloth1)
I already got the answer, but it involved some guessing, what would be the best way to solve this (qb)
Before you do anything else, do you know what 8^{-1/3} actually is? If you can work that out you'll make things considerably simpler for yourself.
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Kvothe the Arcane
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(Original post by SidTheSloth1)
I already got the answer, but it involved some guessing, what would be the best way to solve this (qb)

(Original post by SidTheSloth1)
So I would write 2^2n^3/2 = 2^3/8 ? I would understand better if someone wrote out each step
You could do that. But I made the question more complicated. It'd be simpler, as davros suggests, to just evaluate 8^{-1/3} and take it from there.
Still, I suppose you could say that as 4n^{\frac{3}{2}}=8^{- \frac{1}{3}} then 2^2 \times n^{\frac{3}{2}}=2^{-1}.
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SidTheSloth1
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I did do what davros suggested to get the answer, but I feel that it was mostly guessing, also on the equation you posted, after you fer to that stage, then what do you do?. would you make n the subject by diving by 2^2?
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Kvothe the Arcane
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(Original post by SidTheSloth1)
I did do what davros suggested to get the answer, but I feel that it was mostly guessing, also on the equation you posted, after you fer to that stage, then what do you do?. would you make n the subject by diving by 2^2?
You can do that, yea . Or consider it multiplying the equation by 2^{-2}.
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davros
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(Original post by SidTheSloth1)
I did do what davros suggested to get the answer, but I feel that it was mostly guessing, also on the equation you posted, after you fer to that stage, then what do you do?. would you make n the subject by diving by 2^2?
It's not really "guessing" because
(a) you should be able to recognize powers of 2
(b) in an exam you're probably going to be given something like this where the numbers work out conveniently, rather than some random numbers with roots that you cannot evaluate more easily.

If you want something more systematic, then you should be looking at writing both 4 and 8 as powers of 2, and then setting up an equation that lets you write n as a power of 2 also.
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SidTheSloth1
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Okay, thanks for the help, also I prefer a systematic method because it makes me understand it better
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