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    (Original post by raheem94)
    For question 26,
     \displaystyle 2^{3x+1}=4^{x}
    Take log of both sides,
     \displaystyle log\left(2^{3x+1}\right)=log \left(4^{x}\right) \implies (3x+1)log2=xlog4
    So, (3x+1)log2 = xlog4

    (3x+1) =xlog4/log2

    3x+1=2

    3x = 1

    x = 1/3

    ??? which isn't right
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    (Original post by raheem94)
     \displaystyle log\left(\frac{2x^2}{x^3}\right) \implies log\left(\frac2{x}\right)
    So what does x = 2?
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    (Original post by King-Panther)
    So, (3x+1)log2 = xlog4

    (3x+1) =xlog4/log2

    3x+1=2

    3x = 1

    x = 1/3

    ??? which isn't right
    (3x+1) =xlog4/log2
    3x+1=2x

    You missed out the 'x'.
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    (Original post by King-Panther)
    So what does x = 2?
    According to the question,
     \displaystyle log\left(\frac2{x}\right) = log0.5

    Remove logs,
     \displaystyle \frac2{x}= 0.5
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    (Original post by just george)
    24. yes it is true, the first wiki image steve2005 put up states that if  f''(x)<0 then f has a local maximum at x.
    i.e. if the second derivative of f(x) is less than 0 (negative), there is a local maximum.
    So clearly all local maximum points will produce a negative second derivative
    Consider f(x) = -x^{64}.
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    (Original post by raheem94)
    According to the question,
     \displaystyle log\left(\frac2{x}\right) = log0.5

    Remove logs,
     \displaystyle \frac2{x}= 0.5
    therefore x =2/0.5

    x = 1???

    I take I substitute value of x into the equation to get log(0.5)
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    (Original post by King-Panther)
    therefore x =2/0.5

    x = 1???

    I take I substitute value of x into the equation to get log(0.5)
    2/0.5=4 not 1
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    (Original post by raheem94)
    2/0.5=4 not 1
    2/x = 0.5, if you make x the subject, you multiply 2 by .5 ....


    When I sub 4 into the equation logx+log2x+logx^3 i dont get 0.5
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    (Original post by King-Panther)
    2/x = 0.5, if you make x the subject, you multiply 2 by .5 ....
    No, you'd multiply through by x, then divide through by 0.5, so it does equal 4.
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    (Original post by Brit_Miller)
    No, you'd multiply through by x, then divide through by 0.5, so it does equal 4.
    of course, however, when i sub 4 into the equation i dont get 0.5
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    (Original post by Brit_Miller)
    No, you'd multiply through by x, then divide through by 0.5, so it does equal 4.
    for Q)31, i got 8
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    (Original post by King-Panther)
    2/x = 0.5, if you make x the subject, you multiply 2 by .5 ....


    When I sub 4 into the equation logx+log2x+logx^3 i dont get 0.5
     \displaystyle \frac{2}{x} = 0.5 \implies \frac{2}{x} = \frac12 \implies x \times \frac2{x} = x \times \frac12 \implies 2 = \frac{x}{2} \implies 2 \times 2 = 2\times \frac{x}2 \implies 4=x

    Your equation was,  \displaystyle logx+log2x-logx^3 =log0.5
     \displaystyle log0.5=-0.301...
     \displaystyle log4+log2(4)-log4^3=log4+log8-log64=-0.301...
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    (Original post by Zuzuzu)
    Consider f(x) = -x^{64}.
     f'(x)=-64x^6^3

    so turning point at x=0

     f''(x)=-4032x^6^2

    so at x=0, f''(x)=0

    so not a maximum..?

    i dont understand what you're trying to say? is what i said incorrect? :L
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    (Original post by raheem94)
     \displaystyle \frac{2}{x} = 0.5 \implies \frac{2}{x} = \frac12 \implies x \times \frac2{x} = x \times \frac12 \implies 2 = \frac{x}{2} \implies 2 \times 2 = 2\times \frac{x}2 \implies 4=x

    Your equation was,  \displaystyle logx+log2x-logx^3 =log0.5
     \displaystyle log0.5=-0.301...
     \displaystyle log4+log2(4)-log4^3=log4+log8-log64=-0.301...
    ahhhh

    :hugs:
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    (Original post by raheem94)
     \displaystyle \frac{2}{x} = 0.5 \implies \frac{2}{x} = \frac12 \implies x \times \frac2{x} = x \times \frac12 \implies 2 = \frac{x}{2} \implies 2 \times 2 = 2\times \frac{x}2 \implies 4=x

    Your equation was,  \displaystyle logx+log2x-logx^3 =log0.5
     \displaystyle log0.5=-0.301...
     \displaystyle log4+log2(4)-log4^3=log4+log8-log64=-0.301...
    Q)31 i got 8

    also, for 33, what do i do when 3/2x^2, do i multiply 3 by 2x^2 do its 6x^-2
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    and for (26) couldnt you just do?

     2^{3x+1} = 4^x

     2^{3x+1} = (2\times2)^x

     2^{3x+1} = 2^x\times2^x

     2^{3x+1} = 2^{2x}

     3x+1 = 2x

     x = -1

    Guess if your happy with laws of logs and things either way is fine, dont know why i chose to do it this way
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    (Original post by just george)
    and for (26) couldnt you just do?

     2^{3x+1} = 4^x

     2^{3x+1} = (2\times2)^x

     2^{3x+1} = 2^x\times2^x

     2^{3x+1} = 2^{2x}

     3x+1 = 2x

     x = -1

    Guess if your happy with laws of logs and things either way is fine, dont know why i chose to do it this way
    indeed i could have, thanks.... for 31 is it 8?
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    (Original post by King-Panther)
    indeed i could have, thanks.... for 31 is it 8?
    yes, whats the relevance of your answer though?

    edit:
    (Original post by King-Panther)
    also, for 33, what do i do when 3/2x^2, do i multiply 3 by 2x^2 do its 6x^-2
    what do you mean? you have  \frac{3}{2}x^2 , so to differentiate it you times it by the power, and then take one off the power?
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    (Original post by just george)
    yes, whats the relevance of your answer though?
    Dy/dx tells us what the gradient is when we at x when we sub it in, d2y/dx2 tells us in which direction its going, its a positive number so its a minimum going in the positive direction?
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    (Original post by just george)
    yes, whats the relevance of your answer though?

    edit:


    what do you mean? you have  \frac{3}{2}x^2 , so to differentiate it you times it by the power, and then take one off the power?
    No, the fraction is  \frac{3}{2x^2}

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