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Solving Equations

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    The following has caught me out;

    2x = 512

    I have no idea where to start, but I think the use of logs is needed?

    Could someone point me in the right direction?

    Rep for useful response!
  2. Offline

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    (Original post by MrBlackwood)
    The following has caught me out;

    2x = 512

    I have no idea where to start, but I think the use of logs is needed?

    Could someone point me in the right direction?

    Rep for useful response!
    Take log base 2 of both sides (or if your calculator doesn't support that - just take regular logs of both sides).

    This is one of the most basic logarithm questions - have you studied logs properly yet?
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    (Original post by hassi94)
    Take log base 2 of both sides (or if your calculator doesn't support that - just take regular logs of both sides).

    This is one of the most basic logarithm questions - have you studied logs properly yet?
    Thank you!!

    And no, haha. We were taught the rules in 3 minutes and expected to be experts with no further teaching
  4. Offline

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    Or if you have a calculator...you can simply divide 512 by 2 for 9 times...

    hence you get 2^9 = 512...

    (but taking log I think is the better option)
  5. Offline

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    Loga^x = xLoga

    Hopefully this is a nudge in the right direction :P
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    could also take natural logs...
  7. Offline

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    (Original post by Hasufel)
    could also take natural logs...
    The base is irrelevant as long as it is the same on both sides, just happens that in this case base 2 is convenient for the 512...

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