louise18152
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#1
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Does anyone know how to solve this
3/(e ^ 5) * 2/(root(3, e))
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mqb2766
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#2
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(Original post by louise18152)
Does anyone know how to solve this
3/(e ^ 5) * 2/(root(3, e))
Do you mean simplify?
If so, which bit is causing problems?
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louise18152
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Thanks you but I've figured that one out
However could you help with this one
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laurawatt
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(Original post by louise18152)
Thanks you but I've figured that one out
However could you help with this one
Do you know how to write √(f^7) in index form?
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louise18152
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(Original post by laurawatt)
Do you know how to write √(f^7) in index form?
F^7/2
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mqb2766
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(Original post by louise18152)
F^7/2
Yes but it's on the denominator so what do you o to the power?
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louise18152
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(Original post by mqb2766)
Yes but it's on the denominator so what do you o to the power?
3 1/2
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mqb2766
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(Original post by louise18152)
3 1/2
7/2 is 3 1/2 but
1/(y^(7/2)) = y^?
What happens to the index when you bring it to the numerator?
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louise18152
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Doesn't it go Negative
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mqb2766
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(Original post by louise18152)
Doesn't it go Negative
Yes, so what have you got so far?
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louise18152
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#11
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(Original post by mqb2766)
Yes, so what have you got so far?
So would it be 3
3f-1/2
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mqb2766
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(Original post by louise18152)
So would it be 3
3f-1/2
No, what steps did you do?
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laurawatt
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If I told you 5/(x^2) is the same as 5(x^-2), could you apply the same principle to your Q?
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louise18152
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(Original post by mqb2766)
No, what steps did you do?
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mqb2766
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(Original post by louise18152)
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Not sure what happened to that 3 on the denominator. Without the division of f^5 you have
3/sqrt(f^7)
3 * (1/f^7/2))
3 " f^(-7/2))
That was pretty much what you described in the preceding posts, but your notation went wrong.

Then it's a case of doing something similar and dividing that expression by f^5. Note, using index notation, you could combine fs then negate the power (divide).
Last edited by mqb2766; 4 weeks ago
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