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Help on these Surds and Indices questions desperately needed! watch

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    The image I have attached is on 2 questions (7 and 8) regarding Surds and Indices. I found the previous ones really easy but I am really struggling with the last 2. Would anyone mind helping me with these and explaining them? I would really appreciate it!
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    (Original post by dp00)
    The image I have attached is on 2 questions (7 and 8) regarding Surds and Indices. I found the previous ones really easy but I am really struggling with the last 2. Would anyone mind helping me with these and explaining them? I would really appreciate it!
    What have you tried? What are your thoughts?
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    (Original post by dp00)
    The image I have attached is on 2 questions (7 and 8) regarding Surds and Indices. I found the previous ones really easy but I am really struggling with the last 2. Would anyone mind helping me with these and explaining them? I would really appreciate it!
    7 - Rationalise the denominator by multiplying top and bottom by \sqrt{2}. Proceed to solve for x.

    8 - Rationalise the denominator by multiplying top and bottom by the conjugate of \sqrt{12}-\sqrt{8} (of course, you can simplify these surds too)
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    Q7 =5 root 2
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    I am really sorry, but I just don't understand how you got that answer! Would you mind explaining it and going through the working out?
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    (Original post by dp00)
    I am really sorry, but I just don't understand how you got that answer! Would you mind explaining it and going through the working out?
    No. What have you tried? What are your thoughts, we're here to help you and point you in the right direction, not spoon you the answers. Have you looked at RDK's post? Did you try what he said? What did you get stuck on?
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    (Original post by Zacken)
    No. What have you tried? What are your thoughts, we're here to help you and point you in the right direction, not spoon you the answers. Have you looked at RDK's post? Did you try what he said? What did you get stuck on?
    I am on Q7. I understand that you have to rationalise the denominator, so you multiply 6x by sqrt2 and sqr2 by sqrt2. So you end up with 6xsqrt2 as the numerator and 2 as the denominator. I do not know what to do next...
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    (Original post by dp00)
    I am on Q7. I understand that you have to rationalise the denominator, so you multiply 6x by sqrt2 and sqr2 by sqrt2. So you end up with 6xsqrt2 as the numerator and 2 as the denominator. I do not know what to do next...
    Well, you could do that. But it'd be easier to do this:

    \frac{a}{b} = c \iff a = bc.

    So here, you'd have \displaystyle \frac{6x}{\sqrt{2}} = 10 + x\sqrt{8} which you can re-arrange to:

    \displaystyle 6x = \sqrt{2}(10 + x\sqrt{8}) \iff 6x = 10\sqrt{2} + x\sqrt{8}\sqrt{2} = 10\sqrt{2} + x\sqrt{16}
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Size:  6.6 KBCan someone please tell me how to do this, I keep getting 0.718 or something but the answers .736. its core 3 btw. And I know how to get to the .736 but why wouldn't my way work?
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    (Original post by Andrew.G)

    Sorry for poor quality
    http://www.thestudentroom.co.uk/show...9#post64637319

    Let's not post full solutions to a question. Thanks!
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    (Original post by OGFakiie)
    Name:  math.png
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Size:  6.6 KBCan someone please tell me how to do this, I keep getting 0.718 or something but the answers .736. its core 3 btw. And I know how to get to the .736 but why wouldn't my way work?
    You should make your own thread, not hijack this one, but:

    2\ln x + 1 = \ln 2x \Rightarrow \ln x^2 + \ln e = \ln 2x \Rightarrow \cdots \Rightarrow x = \frac{2}{e}
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    (Original post by OGFakiie)
    Name:  math.png
Views: 77
Size:  6.6 KBCan someone please tell me how to do this, I keep getting 0.718 or something but the answers .736. its core 3 btw. And I know how to get to the .736 but why wouldn't my way work?
    What have you tried to do??
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    (Original post by Andrew.G)
    Attachment 580770

    Sorry for poor quality
    Thank you very much! That is a really easy method! Could you help me with Q8. For Q8, would you rationalise by multiplying by (sqrt12 +sqrt8 divided by sqrt12+sqrt8)?
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    (Original post by RDKGames)
    http://www.thestudentroom.co.uk/show...9#post64637319

    Let's not post full solutions to a question. Thanks!
    Oh sorry, won't happen again
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    okay yeah answer is 2e^-1 , but I got 2-e somehow but all my working looks perfectly fine (to me), where have I gone wrong?
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    (Original post by OGFakiie)
    okay yeah answer is 2e^-1 , but I got 2-e somehow but all my working looks perfectly fine (to me), where have I gone wrong?
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    \ln x + 1 = \ln 2 deson't mean x+e = 2. It means \ln ex = 2.

    Because \ln x + 1 = \ln 2 \iff \ln x + \ln e = \ln 2 \iff \ln ex = \ln 2.

    It's \ln a + \ln b = \ln ab not \ln a + \ln b = \ln (a+b).
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    (Original post by dp00)
    Thank you very much! That is a really easy method! Could you help me with Q8. For Q8, would you rationalise by multiplying by (sqrt12 +sqrt8 divided by sqrt12+sqrt8)?
    Yes you would!

    But first you want to solve the square root of 12 and 8, it will make it easier for!
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    (Original post by Zacken)
    Well, you could do that. But it'd be easier to do this:

    \frac{a}{b} = c \iff a = bc.

    So here, you'd have \displaystyle \frac{6x}{\sqrt{2}} = 10 + x\sqrt{8} which you can re-arrange to:

    \displaystyle 6x = \sqrt{2}(10 + x\sqrt{8}) \iff 6x = 10\sqrt{2} + x\sqrt{8}\sqrt{2} = 10\sqrt{2} + x\sqrt{16}
    i think I have worked out q8. Is it sqrt3 + sqrt2??
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    (Original post by dp00)
    i think I have worked out q8. Is it sqrt3 + sqrt2??
    Yes.
 
 
 
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