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Math Help - Two problems- one with square root, one with powers.

  1. #1
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    Two problems- one with square root, one with powers.



    I'm so confused on these two problems. If you could help at all I would really appreciate it. Thanks!
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  2. #2
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    Quote Originally Posted by lucyluu View Post


    I'm so confused on these two problems. If you could help at all I would really appreciate it. Thanks!
    your question doesn't belong here.
    but anyway,
    \sqrt{n^2+n^2+n^2+n^2} =\sqrt{4n^2}=\sqrt{(2n)^2} =\left | 2n \right |=2n=64
    therefore n=32
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  3. #3
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    Hi there, in the first question, are they addition signs or division signs underneath the square root?

    In the second quesiton use some logs.

     2^x = 15 \Rightarrow x = \log_215

     15^y = 32 \Rightarrow y = \log_{15}32

     x\times y = \log_215 \times \log_{15}32=\dots
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    Quote Originally Posted by pickslides View Post
    Hi there, in the first question, are they addition signs or division signs underneath the square root?
    if they were division signs,there would be no longer n to find
    Last edited by Raoh; December 13th 2009 at 12:44 PM.
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  5. #5
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    Quote Originally Posted by pickslides View Post
    Hi there, in the first question, are they addition signs or division signs underneath the square root?

    In the second quesiton use some logs.

     2^x = 15 \Rightarrow x = \log_215

     15^y = 32 \Rightarrow y = \log_{15}32

     x\times y = \log_215 \times \log_{15}32=\dots
    Thank you for the response. However, my teacher hasn't taught us anything about logs yet... is there any other way to do it?
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  6. #6
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    2^x = 15, \ 15^y = 32 \Rightarrow 15^y = (2^x)^y = 32 = 2^5
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    You can find some estimates by trial and error.

    Consider 2^x = 15

    We know that 2^3 = 8 and 2^4 = 16 as 15 is between 8 and 16 then x must be between 3 and 4, probably closer to 4.

    It now depends on how accuarate you want to be.

    Do the same for 15^y = 32, the answer should be between 1 and 2.
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  8. #8
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    it seems that he only wanted the value of xy which is 5 (as "Defunkt" showed).
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  9. #9
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    \sqrt{n^2 + n^2 + n^2 + n^2} = 64

    \sqrt{4n^2} = 64

    2 \sqrt{n^2} = 64

    \sqrt{n^2} = 32

    Since n > 0, \sqrt{n^2} = n

    So, n = 32.

    --------------------------

    2^x = 15, so x = log_2(15)

    15^y = 32, so y = log_{15}(32)

    Thus xy = log_2(15) \times log_{15}(32).

    Now, you know from the change of base formula that :

    log_2(15) = \frac{ln(15)}{ln(2)}

    log_{15}(32) = \frac{ln(32)}{ln(15)}

    So, xy = \frac{ln(15)}{ln(2)} \times \frac{ln(32)}{ln(15)}

    xy = \frac{ln(15) \times ln(32)}{ln(2) \times ln(15)}

    xy = \frac{ln(32)}{ln(2)} = 5

    Last edited by Bacterius; December 13th 2009 at 01:15 PM.
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