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    Hi all can anyone help me understand how to tackle questions 6 onward in the below paper please? I am really struggling to get started on these, and I thought it was going so well up until then!

    http://www.mrbartonmaths.com/resourc...une%202015.pdf

    The mark scheme is below if it helps!

    http://www.mrbartonmaths.com/resourc...k%20Scheme.pdf

    I would be very grateful for any advice.
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    (Original post by MLW12345)
    Hi all can anyone help me understand how to tackle questions 6 onward in the below paper please? I am really struggling to get started on these, and I thought it was going so well up until then!

    http://www.mrbartonmaths.com/resourc...une%202015.pdf

    The mark scheme is below if it helps!

    http://www.mrbartonmaths.com/resourc...k%20Scheme.pdf

    I would be very grateful for any advice.
    Start with 6a:

    Do you have any ideas for this? Please post your thoughts and working attempts.
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    Hello thank you so much for the reply, I struggle being self taught with no official tutor! Apologies for lack of specifics it's 6b am struggling with as I cannot see how to apply Cos^2x + Sin^2x = 1
    Any pointers/nudges would be fantastic!
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    (Original post by MLW12345)
    Hello thank you so much for the reply, I struggle being self taught with no official tutor! Apologies for lack of specifics it's 6b am struggling with as I cannot see how to apply Cos^2x + Sin^2x = 1
    Any pointers/nudges would be fantastic!
    \cos^2 \theta+\sin^2 \theta=1

    \therefore \sin^2 \theta = {?}
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    So all I had to do was replace sin^2x with 1-cos^x?? dear me I was trying to move the sin term to the left….Thank you so much, phew! If for a couple more minutes please could you also help me with the last parts of this question?

    I managed to get: LogaK - Loga2 = 2/3

    into LogaK/2 = 2/3

    then inv logs to get A^2/3 = K/2 but am not sure how to get A on its own according to mark scheme?

    The last question I cannot even get started on! I have attached it but am in a bit of a pickle
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    (Original post by MLW12345)
    ...
    What happens if you raise both sides to the power \frac{3}{2}?
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    (Original post by Zacken)
    What happens if you raise both sides to the power \frac{3}{2}?
    Hello thank you so much for getting back, that is exactly what the mark scheme says to do, but it wasn't natural for me to think like that. I'd want to undo the index with some form of rooting.
    Please can you tell me why I need to raise to power 3/2?
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    (Original post by MLW12345)
    Hello thank you so much for getting back, that is exactly what the mark scheme says to do, but it wasn't natural for me to think like that. I'd want to undo the index with some form of rooting.
    Please can you tell me why I need to raise to power 2/3?
    If you have an equation of the form a^n = k then so find a you need to raise both sides to the power \frac{1}{n} so that you get (a^{n})^{1/n} = k^{1/n} \Rightarrow a^{n/n} = k^{1/n} \Rightarrow a = k^{1/n}.

    In your case we have a^{2/3} = k \Rightarrow (a^{2/3})^{3/2} = k^{3/2} \Rightarrow a^{\frac{3}{2} \times \frac{2}{3}} = k^{3/2} \Rightarrow a^1 = k^{3/2} \Rightarrow a = k^{3/2}
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    (Original post by Zacken)
    If you have an equation of the form a^n = k then so find a you need to raise both sides to the power \frac{1}{n} so that you get (a^{n})^{1/n} = k^{1/n} \Rightarrow a^{n/n} = k^{1/n} \Rightarrow a = k^{1/n}.

    In your case we have a^{2/3} = k \Rightarrow (a^{2/3})^{3/2} = k^{3/2} \Rightarrow a^{\frac{3}{2} \times \frac{2}{3}} = k^{2/3} \Rightarrow a^1 = k^{2/3} \Rightarrow a = k^{2/3}
    Hi Zacken thanks again. Up until now I have never had to use this rule so thank you for putting me in the right direction! I am going back to looking at this section as of now
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    (Original post by MLW12345)
    Hello thank you so much for getting back, that is exactly what the mark scheme says to do, but it wasn't natural for me to think like that. I'd want to undo the index with some form of rooting.
    Please can you tell me why I need to raise to power 2/3?
    You may prefer to think of it like this..

    \displaystyle A^{\frac{2}{3}} : this means the cube root of A squared.

    So you can start by cubing both sides

    \displaystyle A^{\frac{2}{3}} = \frac{k}{2}

    \displaystyle \Rightarrow A^2 = \left(\frac{k}{2}\right)^3 = \frac{k^3}{8}

    Then square root both sides.
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    (Original post by notnek)
    You may prefer to think of it like this..

    \displaystyle A^{\frac{2}{3}} : this means the cube root of A squared.

    So you can start by cubing both sides

    \displaystyle A^{\frac{2}{3}} = \frac{k}{2}

    \displaystyle \Rightarrow A^2 = \left(\frac{k}{2}\right)^3 = \frac{k^3}{8}

    Then square root both sides.

    Absolute winner! I tried this to start off with but I didn't see it in the mark scheme so binned it off!
    I've been playing around with both reciprocal index and rooted forms since being shown this thank you!
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    Hi all, thank you for every single response I am already much more confident!

    I attached the last question worth 5 marks below and I attempted it using what I know about logs, I expanded the brackets and used the multiplication log law for the other side.
    what I got was true to the mark scheme and I carried on to factor and solve but am not really clear how the logs disappear when they are equal to each other?

    Before factoring I ended up with Log2(left equation) = Log2 (right equation) I just moved everything onto one side, factored and solved?

    Please can I have any explanation as to what is actually happening that makes the logs disappear?
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    (Original post by MLW12345)
    ...
    If you have \log_a x = \log_a y then it must be the case that x=y since logs (to the same base) are only equal if their 'arguments' are equal.

    You're saying that taking the logarithm of a number x is the same as the logarithm of the number y and this only occurs when x and y are the same numbers. Here x and y are expressions, so you can equate them.

    The reason this works for logarithms is because the logarithm is an injective function (fancy terminology and you don't need to know it). The above should suffice.
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    (Original post by Zacken)
    If you have \log_a x = \log_a y then it must be the case that x=y since logs (to the same base) are only equal if their 'arguments' are equal.

    You're saying that taking the logarithm of a number x is the same as the logarithm of the number y and this only occurs when x and y are the same numbers. Here x and y are expressions, so you can equate them.

    The reason this works for logarithms is because the logarithm is an injective function (fancy terminology and you don't need to know it). The above should suffice.

    Legend! I was hoping it was some thing of the case so just carried on but your explanation perfectly satisfies.I am very clear now thank you!
 
 
 
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