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C1 indices

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    How would you find x?...

    4^x=1/4

    thanks
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    (Original post by jessica_anne_clu)
    How would you find x?...

    4^x=1/4

    thanks
    ln both sides if you want to show it mathmatically.
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    (Original post by jessica_anne_clu)
    How would you find x?...

    4^x=1/4

    thanks
    e.g.

     2^{2x} = \frac12 

2^{2x} = 2^{-1}

2x = -1

x = -\dfrac12
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    (Original post by jessica_anne_clu)
    How would you find x?...

    4^x=1/4

    thanks
    What power would you use to get the reciprocal of a number?
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    (Original post by roar558)
    ln both sides if you want to show it mathmatically.
    but how would you go about it? I know the answer is -1 but I don't know how you'd get it
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    In this case, try multiplying both sides by 4:

    4^{x+1}=1

    Can you see the answer now?


    A general approach for any question like this involves logarithms, but I don't think they're covered in C1...?
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    (Original post by jessica_anne_clu)
    but how would you go about it? I know the answer is -1 but I don't know how you'd get it
    Just write down the answer. No working required.
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    (Original post by roar558)
    ln both sides if you want to show it mathmatically.
    ln is in C3, while OP is studying C1 so she probably doesn't knows logarithms.
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    You could use logarithms:
    Log 4^x = Log 0.25
    x Log 4 = Log 0.25
    x= (Log 0.25)/(Log 4)
    x=-1
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    (Original post by jessica_anne_clu)
    but how would you go about it? I know the answer is -1 but I don't know how you'd get it
     \dfrac12 = 2^{-1}

\dfrac14 = ?
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    (Original post by raheem94)
    e.g.

     2^{2x} = \frac12 

2^{2x} = 2^{-1}

2x = -1

x = -\dfrac12
    x = -1, as

    4x = \frac{1}{4}
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    (Original post by tnetennba)
    You could use logarithms:
    Log 4^x = Log 0.25
    x Log 4 = Log 0.25
    x= (Log 0.25)/(Log 4)
    x=-1
    OP is studying C1 indices, so she probably doesn't knows logarithms.
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    (Original post by thegodofgod)
    x = -1, as

    4x = \frac{1}{4}
    I didn't wanted to give the solution hence i was giving a similar example, i know the answer for OP's question is -1.
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    (Original post by james.h)
    In this case, try multiplying both sides by 4:

    4^{x+1}=1

    Can you see the answer now?


    A general approach for any question like this involves logarithms, but I don't think they're covered in C1...?
    4^{x}=\frac{1}{4}

    2^{2x}=\frac{1}{4}

    2^{2x}=[\frac{1}{2}]^2

    2^{2x}=2^{-1 \times 2}

    2^{2x}=2^{-2}

    2x=-2

    x=-1
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    (Original post by raheem94)
    OP is studying C1 indices, so she probably doesn't knows logarithms.
    You're right I am studying C2 at the moment but I am retaking C1
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    (Original post by raheem94)
    I didn't wanted to give the solution hence i was giving a similar example, i know the answer for OP's question is -1.
    My bad
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    (Original post by thegodofgod)
    4^{x}=\frac{1}{4}

    2^{2x}=\frac{1}{4}

    2^{2x}=[\frac{1}{2}]^2

    2^{2x}=2^{-1 \times 2}

    2^{2x}=2^{-2}

    2x=-2

    x=-1
    Isn't it easier to do it in the below way:
     4^x = \dfrac14

4^x = 4^{-1}

x=-1
    Involves only 2 steps.
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    (Original post by thegodofgod)
    ...full solution...
    I did say "for any question". Try that method you've stated on something like a^x = 1/b :p:

    I get your point, though. :yep:
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    (Original post by raheem94)
    Isn't it easier to do it in the below way:
     4^x = \dfrac14

4^x = 4^{-1}

x=-1
    Involves only 2 steps.
    Hmm - didn't even notice that :giggle:

    Think it was a good decision of mine to drop maths after AS
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    Why would you convert everything to powers of 2 when it's already powers of 4? :/

    All you do for this question is rewrite 1/4 as 4^(-1) and compare indices.

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