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    Hey guys,

    I'm a bit stuck with part 3 of this question. Mainly with rearranging the equation to get a half somewhere. I worked out the answer to the first half using a different method but I can't quite get the method they're using to get the answer to the y coordinate of p

    x = -0.32

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    (Original post by Saywhatyoumean)
    Hey guys,

    I'm a bit stuck with part 3 of this question, rearranging the equations to get a half in the equation. I worked out the answer to the first half of using a different method but I can't quite get the method they're using to get the answer to the y coordinate of p

    x = -0.32

    Name:  Screen Shot 2016-02-28 at 23.12.45.png
Views: 88
Size:  389.4 KB Attachment 508687508689

    please post your workings
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    (Original post by TeeEm)
    please post your workings
    Okay well I put the two equations equal to each other straight away and took logs of both sides:

    (1/3)^x = 2(3^x)

    Xlog1/3 = log 2(3^x)

    xlog(1/3) = log2 + xlog3

    Xlog1/3 - xlog3 = log2

    X(log1/3 - log3) = log2

    X= log2/log1/9

    X= -0.32


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    (Original post by Saywhatyoumean)
    ...
    If you've found x, then you can just plug that value right back any of your two equations y=  \left(\frac{1}{3}\right)^x or y = 2(3^x) and they'll both give you the y-coordinate out.

    It's kind of like how, if you have a line y = 2x +3 and y = x + 4 then they intersect at 2x + 3 = x + 4 \iff x = 1, which is the x-coordinate of the intersection, but also the y-coordinate is given by plugging that into y = 2x + 3 = 2(1) +3 = 5 or y = x + 4 = 1 +4 = 5.
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    (Original post by Saywhatyoumean)
    Okay well I put the two equations equal to each other straight away and took logs of both sides:

    (1/3)^x = 2(3^x)

    Xlog1/3 = log 2(3^x)

    xlog(1/3) = log2 + xlog3

    Xlog1/3 - xlog3 = log2

    X(log1/3 - log3) = log2

    X= log2/log1/9

    X= -0.32


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    (Original post by Zacken)
    If you've found x, then you can just plug that value right back any of your two equations y=  \let(\frac{1}{3}\right)^x or y = 2(3^x) and they'll both give you the y-coordinate out.

    It's kind of like how, if you have a line y = 2x +3 and y = x + 4 then they intersect at 2x + 3 = x + 4 \iff x = 1, which is the x-coordinate of the intersection, but also the y-coordinate is given by plugging that into y = 2x + 3 = 2(1) +3 = 5 or y = x + 4 = 1 +4 = 5.
    Zacken has already started helping you
    all the best
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    (Original post by Zacken)
    If you've found x, then you can just plug that value right back any of your two equations y=  \let(\frac{1}{3}\right)^x or y = 2(3^x) and they'll both give you the y-coordinate out.

    It's kind of like how, if you have a line y = 2x +3 and y = x + 4 then they intersect at 2x + 3 = x + 4 \iff x = 1, which is the x-coordinate of the intersection, but also the y-coordinate is given by plugging that into y = 2x + 3 = 2(1) +3 = 5 or y = x + 4 = 1 +4 = 5.
    Thanks for this, I knew that but x was a very long decimal and subbing it in to get y didn't give me a nice answer of root 2, it was a few decimal places out so I didn't think I'd get the marks for doing that here? I'm not sure though


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    (null)

    Again thanks, I got roughly something like that but what happens to the 1^x in that - wouldn't it be 1^x/2 = 3^x ?

    I feel like that's a stupid question


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    (Original post by TeeEm)
    Zacken has already started helping you
    all the best
    Thanks lol


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    (Original post by Saywhatyoumean)
    Thanks for this, I knew that but x was a very long decimal and subbing it in to get y didn't give me a nice answer of root 2, it was a few decimal places out so I didn't think I'd get the marks for doing that here? I'm not sure though


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    You're not meant to use the decimal answer, you should use the exact value instead:

    So you know that x = \displaystyle \frac{\log 2}{\log \frac{1}{9}}, then:

    How can you work out 3^x exactly? Use the fact that 3^{\log_3 m} =m

    Does that help?
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    (Original post by Zacken)
    You're not meant to use the decimal answer, you should use the exact value instead:

    So you know that x = \displaystyle \frac{\log 2}{\log \frac{1}{9}}, then:

    How can you work out 3^x exactly? Use the fact that 3^{\log_3 m} =m

    Does that help?
    Ohh yes it does! Still would never have thought to do that in exam - but what about with this method:

    So if (1/3)^x = (1^x/3^x) = 2(3^x)

    1^x = 2 x (3^2x)

    1^x / 2 = 3^2x ?

    What happens to the 1^x ?




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    (Original post by Saywhatyoumean)
    Ohh yes it does! Still would never have thought to do that in exam - but what about with this method:

    So if (1/3)^x = (1^x/3^x) = 2(3^x)

    1^x = 2 x (3^2x)

    1^x / 2 = 3^2x ?

    What happens to the 1^x ?




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    1^x = 1 for all x.
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    Anyone know which paper this is from?
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    (Original post by Zacken)
    1^x = 1 for all x.
    omg of course how did I not see that😂😂thanks!


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    (Original post by anonwinner)
    Anyone know which paper this is from?
    it's one of the OCR Solomon papers, H I think


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    (Original post by Saywhatyoumean)
    it's one of the OCR Solomon papers, H I think


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    thanks
 
 
 
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