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C1 indices

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  1. Offline

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    How would you find x?...

    4^x=1/4

    thanks
  2. Offline

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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
  4. Offline

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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?
  5. Offline

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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
  6. Offline

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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...?
  7. Offline

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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.
  8. Offline

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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.
  9. Offline

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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.
  14. Offline

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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
  15. Offline

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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
  16. Offline

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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
  17. Offline

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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.
  18. Offline

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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:
  19. Offline

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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
  20. Offline

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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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