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Math Help - solve the following

  1. #1
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    solve the following

    kin the equation solve the following

    16^k=2^2k+6



    my classmate tells me the answer is k =3, is this right ?

    Muchos gracias
    Last edited by Greg; January 28th 2007 at 08:33 AM.
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  2. #2
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    Hello, Greg!


    k 2k+6
    16 = 2

    my classmate tells me the answer is k =3, is this right ?

    I have no idea . . .

    What is the question?

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  3. #3
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    sorry for the unclear question

    I cut and paste it from word and it came out badly. I think I have solved problem for now
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  4. #4
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    Quote Originally Posted by Greg View Post
    kin the equation solve the following

    16^k=2^2k+6



    my classmate tells me the answer is k =3, is this right ?

    Muchos gracias
    16^k=2^{2k}+6,

    may be written:

    2^{4k}=2^{2k}+6.

    Now put y=2^{2k}, and we have a quadratic in y:

    y^2=y+6

    or:

    y^2-y=6=0

    which has roots y=3, \ -2, but as we are interested in real solutions for k we have y=-2 does not give a real k, so we are left with y=3, then:

    3=2^{2k}

    take logs:

    \log(3)=2\,k \log(2)

    k=\log(3)/(2\log(2)) \approx 0.792

    RonL
    Last edited by CaptainBlack; January 28th 2007 at 12:09 PM. Reason: Incorporate AfterShock's correction
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  5. #5
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    Hello, Greg!

    Solve: . 16^k\:=\:2^{2k+6}

    We have: . (2^4)^k\:=\:2^{2x+6}\quad\Rightarrow\quad 2^{4k}\:=\:2^{2k+6}

    Since the bases are equal, the exponents are equal:

    . . 4k\:=\:2k+6\quad\Rightarrow\quad2k\:=\:6\quad\Righ  tarrow\quad k \:=\:3

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  6. #6
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    Quote Originally Posted by CaptainBlack View Post
    16^k=2^{2k}+6,

    may be written:

    2^{4k}=2^{2k}+6.

    Now put y=2^k, and we have a quadratic in y:

    y^2=y+6

    or:

    y^2-y=6=0

    which has roots y=3, \ -2, but as we are interested in real solutions for k we have y=-2 does not give a real k, so we are left with y=3, then:

    3=2^k

    take logs:

    \log(3)=k \log(2)

    k=\log(3)/\log(2) \approx 1.585

    RonL
    This is the way I thought of this problem too. Would have been nice if Greg used parentheses. Any way, your answer is off by a factor of 1/2. You went wrong during your substitution; answer should be: ln(3)/(2*ln(2)) or .7925. Obviously this is irrelevant since it's not even the question he's asking.
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