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Indices - Core 1

Hi,
I'm having some real problems with some indices questions for a maths examination. Any help would be appreciated.
Regards.
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it is better to go back to the start and cross multiply.
Hope that helps :smile:

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Moved to maths :h:
Original post by OL350
Hi,
I'm having some real problems with some indices questions for a maths examination. Any help would be appreciated.
Regards.

Alternative method: "1 is wrong in one direction, because 2<642 < 64. To fix it, we need to make x bigger. Let's try the next fourth-power, which is 16. Oh look, it works. There is only one answer, because the RHS is decreasing in x while the LHS is increasing. Therefore, the answer is x=16."

Might not get many marks in the exam, but it's perfectly rigorous.
Reply 5
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OL350
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Any ideas on this one?
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Do exactly as I did before. Get the numbers to one side and the variable x to the other side

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Reply 7
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OL350
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I really can't seem to do this. Once X by 4 I'm unsure on how to proceed.
Remember you're adding the indices when you bring x to the left handside. Its x by 3 + x by 0.5 giving you x by 3.5 instead of x by 4

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OL350
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Any chance you could show the answer how you previously have, really not understanding it.
Original post by OL350
Any ideas on this one?


Get rid of 4 by x both sides by 4

rootx = x^1/2

you can put negative powers at the bottom eg) 4x^-3 = 4/x^3

then do the same as above post
Reply 11
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OL350
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Thanks for the help, very much appreciated. As you can see from the attachments I think I've finally got it, but one more question - I have highlighted the stage where I have multiplied by the power of 2/7, may someone please explain why we multiply by the reciprocal (also done in the first example)?
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Original post by OL350
Thanks for the help, very much appreciated. As you can see from the attachments I think I've finally got it, but one more question - I have highlighted the stage where I have multiplied by the power of 2/7, may someone please explain why we multiply by the reciprocal (also done in the first example)?


The person who provided the solution used poor terminology ... You are not multiplying by a power ... You are raising both sides to a power

Consider

ab=ca^b = c

So

(ab)d=cd(a^b)^d = c^d

So

abd=cda^{bd} = c^d

In your questions a is the value that you are trying to find so you want it on its own

For this to happen bd=1 hence the reciprocal
(edited 9 years ago)

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