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    Alright Guys, looking to hit 90ums in c3 and c4 and thought that i'd try and get the laws and rules together which always crop up..

    By these i mean the Indicies, surds, log/ln's (do Logs come up in C3/C4 or is it only Ln?) Etc


    Thanks sorry it was a pretty rubbish explanation..
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    (Original post by Nilsdejongh)
    Alright Guys, looking to hit 90ums in c3 and c4 and thought that i'd try and get the laws and rules together which always crop up..

    By these i mean the Indicies, surds, log/ln's (do Logs come up in C3/C4 or is it only Ln?) Etc


    Thanks sorry it was a pretty rubbish explanation..
    I think logs show up in A2, and laws you'll need:

    Surds:

    / = root sign btw :P

    /a•/b = /a•b
    /a•a•b = a/b
    n/a + m/a = (n+m)/a

    When dividing a number by the square root of itself, it becomes the square root. (eg. dividing 3 by /3 = /3)

    Think that's it for surds except the obvious. (/a•/a = a)

    Indices:

    From now on, / = divide.
    x^n • x^m = x^(n+m)
    x^n / x^m = x^(n-m)
    (x^n)^m = x^nm
    (a+x)^n = b^n(a/b + x/b)^n 1.

    Logs:

    Loga + logb = log(ab)
    Loga - logb = log(a/b)
    Log(a)^n = n•loga
    Logb(b) = 1 {sorry the first b here is supposed to be the base, so like when you normally use the log button on the calculator it's to the base 10, and therefore when you calculate log(10) is equals 1.)

    These obviously all work for natural logs too.

    Oh also e^ln(a) = a and 10^log(a) = a.

    Hope I helped and sorry about the crappy typed equations, on my iPod.


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    (Original post by Nilsdejongh)
    Alright Guys, looking to hit 90ums in c3 and c4 and thought that i'd try and get the laws and rules together which always crop up..

    By these i mean the Indicies, surds, log/ln's (do Logs come up in C3/C4 or is it only Ln?) Etc


    Thanks sorry it was a pretty rubbish explanation..
    The core modules build on the previous ones, so if you're aiming for top marks in C3 and C4 you should know the C1 and C2 content inside out.
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    (Original post by Nilsdejongh)
    Alright Guys, looking to hit 90ums in c3 and c4 and thought that i'd try and get the laws and rules together which always crop up..

    By these i mean the Indicies, surds, log/ln's (do Logs come up in C3/C4 or is it only Ln?) Etc


    Thanks sorry it was a pretty rubbish explanation..
    Al the log rules are exactly the same for ln, so if you know one set of rule then you know the other.
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    Suppose you have something like ln(y) = 2x + A, where A is a constant

    Then y = e^{A}e^{2x} \not= (e^{2x} + e^A)
    Spoiler:
    Show
    This is because you take the exponential of both sides, i.e. e^{ln(y)} = e^{(2x+A)}

    Using a^m a^n = a^{m+n} you get the correct result.


    Simplifying, you get y=Be^{2x} where B=e^A


    Also, ln(x) is just log_e(x). Logs are often taken to the base e, so having a convenient way of writing that is useful.
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    (Original post by F1Addict)
    Suppose you have something like ln(y) = 2x + A, where A is a constant

    Then y = e^{A}e^{2x} \not= (e^{2x} + e^A)
    Spoiler:
    Show
    This is because you take the exponential of both sides, i.e. e^{ln(y)} = e^{(2x+A)}

    Using a^m a^n = a^{m+n} you get the correct result.


    Simplifying, you get y=Be^{2x} where B=e^A


    Also, ln(x) is just log_e(x). Logs are often taken to the base e, so having a convenient way of writing that is useful.
    Ah yes, I forgot all about this one.


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    (Original post by F1Addict)
    Suppose you have something like ln(y) = 2x + A, where A is a constant

    Then y = e^{A}e^{2x} \not= (e^{2x} + e^A)
    Spoiler:
    Show
    This is because you take the exponential of both sides, i.e. e^{ln(y)} = e^{(2x+A)}

    Using a^m a^n = a^{m+n} you get the correct result.


    Simplifying, you get y=Be^{2x} where B=e^A


    Also, ln(x) is just log_e(x). Logs are often taken to the base e, so having a convenient way of writing that is useful.
    Could you show how you simplified to get that?
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    e^(2x+A) = e^A•e^2x. Because A is a constant, it follows that e^A is also constant. This means that you can switch in B as a constant to replace e^A as it just looks tidier and easier to use. (And if you need to find A just do lnB)


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