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    (Original post by kiiten)
    What about this question - im not sure where you would start. Ive attached some workingAttachment 513103513107

    Attachment 513103
    Why don't you let u = \log_3 x and v = \log_3 y? Then solve for v and u and then back-sub to find x and y. If you do it this way, you get:

    \log_3 xy = \log_3 x + \log_3 y = u + v = 5

    And \log_3 x \log_3 y = uv = 6 so you just need to solve u+v = 5 and uv =6 which should be easy (it becomes a quadratic in u or v). Does that help?
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    (Original post by notnek)
    You have
    This is why I need to refresh the page before posting. :lol:
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    (Original post by notnek)
    You have

    \log_3 x + log_3 y = 5

    \log_3 x \times log_3 y = 6


    Let a=\log_3 x and b=\log_3 y then you have

    a+b=5

    ab = 6

    So you can ignore logs and solve these equations for a and b. Then substitute back in for a and b.
    Thanks, when i substitute back in should i change a and b back to logs?
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    (Original post by kiiten)
    Thanks, when i substitute back in should i change a and b back to logs?
    Yes, you should (let's say a=8, I'm not saying it is, but just an example) then you would do \log_3 x = 8 \Rightarrow x = 3^8. Do that for whatever values of a, b you get.
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    (Original post by Zacken)
    This is why I need to refresh the page before posting. :lol:
    I always use 'preview post' before posting to see if anything has been added to the thread
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    (Original post by Zacken)
    Yes, you should (let's say a=8, I'm not saying it is, but just an example) then you would do \log_3 x = 8 \Rightarrow x = 3^8. Do that for whatever values of a, b you get.
    Ok, ive done that and I got 27 and 8 for both x and y (4 solutions). 27 is right but 8 is wrong - its supposed to be 9. Where have I gone wrong?

    u + v = 5 so u = 5 - v
    subbed this into uv = 6 to get v^2 - 5v - 6 =0
    so v=3 v=2 which I put into u = 5 - v to get u=2 and u=3
    log3x = 2 etc. to get 27 and 8 ?
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    (Original post by kiiten)
    Ok, ive done that and I got 27 and 8 for both x and y (4 solutions). 27 is right but 8 is wrong - its supposed to be 9. Where have I gone wrong?

    u + v = 5 so u = 5 - v
    subbed this into uv = 6 to get v^2 - 5v - 6 =0
    so v=3 v=2 which I put into u = 5 - v to get u=2 and u=3
    log3x = 2 etc. to get 27 and 8 ?
    Those are your solutions for x - what about the corresponding ones for y?
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    (Original post by Zacken)
    Those are your solutions for x - what about the corresponding ones for y?
    Sorry I meant x=8, x=27 and y=27, y=8. But, the answer is supposed to be 9 not 8.
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    (Original post by kiiten)
    Sorry I meant x=8, x=27 and y=27, y=8. But, the answer is supposed to be 9 not 8.
    If \log_3 x = 2 then x = 3^2 = 9 not x=2^3 = 8. :-)

    In general: \log_a x = y \Rightarrow x = a^y \neq y^a
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    (Original post by Zacken)
    If \log_3 x = 2 then x = 3^2 = 9 not x=2^3 = 8. :-)

    In general: \log_a x = y \Rightarrow x = a^y \neq y^a
    Ahh I see its another log rule. Thanks
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    Referring back to the point mentioned earlier when logs can cancel out if you had
    log2 (72) = m+nlog2 (3)
    I know n=2 from a previous question so would 72 = 3 ^2+m ?

    I'm not sure I'm doing this right but the question is asking to show that log2 (72) = m+nlog2 (3)
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    (Original post by kiiten)
    Referring back to the point mentioned earlier when logs can cancel out if you had
    log2 (72) = m+nlog2 (3)
    I know n=2 from a previous question so would 72 = 3 ^2+m ?
    No! You can only cancel logs when you have only one log on each side.

    I'm not sure I'm doing this right but the question is asking to show that log2 (72) = m+nlog2 (3)
    Post the question, please.
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    (Original post by Zacken)
    No! You can only cancel logs when you have only one log on each side.



    Post the question, please.
    1.c)
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    I worked out n=2 from the previous question. So if you rearrange would it be 2+m = log2(72) / log2(3)
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    (Original post by kiiten)
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    I worked out n=2 from the previous question. So if you rearrange would it be 2+m = log2(72) / log2(3)
    All they want you to do is write \log_2 72 = (8 \times 9) = \log_2 8 + \log_2 9

    and then you know that \log_2 9 = n \log_2 3 and that \log_2 8 = \log_2 2^3 = 3 \log_2 2 = 3.

    So that you can then write \log_2 72 = m + n\log_2 3 for m, n that you've found.
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    (Original post by Zacken)
    All they want you to do is write \log_2 72 = (8 \times 9) = \log_2 8 + \log_2 9

    and then you know that \log_2 9 = n \log_2 3 and that \log_2 8 = \log_2 2^3 = 3 \log_2 2 = 3.

    So that you can then write \log_2 72 = m + n\log_2 3 for m, n that you've found.
    Hm ok but why do you need the last part 3log2(2)=3
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    (Original post by kiiten)
    Hm ok but why do you need the last part 3log2(2)=3
    Because that's your m.
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    (Original post by Zacken)
    Because that's your m.
    Oh so its mlog2(3) + nlog2(3)
    Thanks
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    (Original post by kiiten)
    Oh so its mlog2(3) + nlog2(3)
    Thanks
    No! It's \log_2 72 = \log_2 (8\times 9) = \log_2 8 + 2\log_2 3 = 3 + 2\log_2 3 so m=3 and n=2.
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    (Original post by Zacken)
    No! It's \log_2 72 = \log_2 (8\times 9) = \log_2 8 + 2\log_2 3 = 3 + 2\log_2 3 so m=3 and n=2.
    Oops so does the original question mean that log2(72) = (m+n)log2(3)
    = 2log2(3) + 3log2(2)
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    if you have to determine if a point is maximum of minimum do you use the first or second derivative? Also, how do you know if it is max or min - is it +ve is max?
 
 
 

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