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    Hi, this question is from the Core 4 Edexcel textbook review exercise (question 38).

    Find the coordinates of the minimum point on the curve with equation y=x2x

    I worked out the x coordinate (-1/ln2) and this is correct according to the answers. Substituting this value into the original y equations, I assumed I could find the coordinate for y, but I seem to be unable to rearrange the equation to get an answer.

    I'm stuck at y= (-1/ln2)(2^(-1/ln2))
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    (Original post by ameliagrrh)
    Hi, this question is from the Core 4 Edexcel textbook review exercise (question 38).

    Find the coordinates of the minimum point on the curve with equation y=x2x

    I worked out the x coordinate (-1/ln2) and this is correct according to the answers. Substituting this value into the original y equations, I assumed I could find the coordinate for y, but I seem to be unable to rearrange the equation to get an answer.

    I'm stuck at y= (-1/ln2)(2^(-1/ln2))
    Rewriting \ 2^x \ as\  e^{x ln2} may help.

    EDIT: Having now attempted this question, I'd suggest that if you wanted to simplify further recall that:
    \ a^{-x} = \frac{1}{a^x}
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    I re-wrote it as y=-1/ln2 x 1/2^(ln2) but I can't seem to work it out, it's not clicking in my mind! :P
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    (Original post by ameliagrrh)
    I re-wrote it as y=-1/ln2 x 1/2^(ln2) but I can't seem to work it out, it's not clicking in my mind! :P
    There's your problem: It should read:
    \ 2^{\frac{1}{ln2}} in the denominator of the second fraction.
    Then read my first post
    Welcome to TSR btw
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    I GOT IT. Oh I love it when things click
    Thanks so much for your help!
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    (Original post by ameliagrrh)
    I GOT IT. Oh I love it when things click
    Thanks so much for your help!
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