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Math Help - My first proofs

  1. #1
    Member integral's Avatar
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    My first proofs

    I had never heard of a proof before coming to this forum (Schools FTW ).
    And I though I would try some for fun.
    The reason I am posting them here is because I believe they are not proofs, but rather a silly play in which nothing is proven.
    (I considered putting this in the 'other' forum, but it contains mostly algebra. Sorry If I was wrong.)

    Prove:

    log_bxy=log_bx+lob_by\,\,\,\because
    if
    log_bx=n
    log_by=z
    then
    b^{z+n}=xy
    and
    z+n=log_bxy
    so
    <br />
b^{log_bxy}=xy
    xy=xy


    Prove:
    log_b\frac{x}{y}=log_bx-log_by\,\,\,\because
    say:
    log_bx=\gamma
    log_by=\delta
    Then
    log_b\frac{x}{y}=\gamma-\delta
    so
    b^{\gamma-\delta}=\frac{x}{y}
    and
    {\gamma-\delta}=log_bx-log_by



    Prove:

    log_b(x^y)=ylog_bx\,\,\,\because
    \frac{log_b(x^y)}{y}=log_bx
    log_b(\sqrt[y]{x^y})=log_bx
    log_bx=log_bx

    Are theses proofs? Or not?
    Last edited by integral; April 5th 2010 at 02:59 AM.
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by integral View Post
    I had never heard of a proof before coming to this forum (Schools FTW ).
    And I though I would try some for fun.
    The reason I am posting them here is because I believe they are not proofs, but rather a silly play in which nothing is proven.
    (I considered putting this in the 'other' forum, but it contains mostly algebra. Sorry If I was wrong.)

    Prove:

    log_bxy=log_bx+lob_by\,\,\,\because
    if
    log_bx=n
    log_by=z
    then
    b^{z+n}=xy
    and
    z+n=log_bxy
    so
    <br />
b^{log_bxy}=xy
    xy=xy


    Prove:
    log_b\frac{x}{y}=log_bx-log_by\,\,\,\because
    say:
    log_bx=\gamma
    log_by=\delta
    Then
    log_b\frac{x}{y}=\gamma-\delta
    so
    b^{\gamma-\delta}=\frac{x}{y}
    and
    {\gamma-\delta}=log_bx-log_by



    Prove:

    log_b(x^y)=ylog_bx\,\,\,\because
    \frac{log_b(x^y)}{y}=log_bx
    log_b(\sqrt[y]{x^y})=log_bx
    log_bx=log_bx

    Are theses proofs? Or not?
    I've only looked at the first one so far, but my impression is that you are almost there, except you're supposed to end with log_bxy=log_bx+log_by and not xy = xy!

    By the way I was unfamiliar with the because symbol before now, neat.

    Here's how I would change your steps:

    Prove that log_bxy=log_bx+log_by.

    Proof:

    Let n=log_bx and z=log_by with b, x, y \in{R}, b > 0, x > 0, y > 0, b\neq1.

    Then b^{n+z}=b^nb^z=xy

    so n+z=log_bxy.

    Replacing n and z on the LHS according to the original equations, we obtain log_bxy=log_bx+log_by.

    (Edited for clarity)

    The second proof would be very similar to the first, but the third is a bit different.

    Prove that log_b(x^y)=ylog_bx.

    Proof:

    Let x=b^z with b, z \in{R}, b > 0, b\neq1.

    Then by definition, z=log_bx.

    Substituting the value for z into the original equation, we have x=b^{log_bx}.

    We can raise each side to the power y \in{R}. (We are relying on the fact that u=v \Longrightarrow f(u)=f(v) for any function.)

    So x^y=(b^{log_bx})^y=b^{y*log_bx}.

    Taking the log base b of both sides, log_b(x^y)=ylog_bx.
    Last edited by undefined; April 5th 2010 at 04:57 AM.
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  3. #3
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    Hello, integral!

    Like "undefined", I have my own approach to these . . .


    Prove: . \log_b(xy) \:=\: \log_b(x) + \log_b(y)


    \text{Let: }\;\begin{Bmatrix}\log_b(x) \:=\:P & [1] \\ \log_b(y) \:=\:Q & [2] \end{Bmatrix}\quad\text{Then: }\;\begin{Bmatrix}x \:=\:b^P & [3] \\ y \:=\:b^Q & [4]\end{Bmatrix}


    Multiply [3] and [4]: . x\cdot y \:=\:b^P\cdot b^Q \quad\Rightarrow\quad xy \:=\:b^{P+Q}


    Take logs, base b\!:\;\;\log_b(xy) \:=\:\log_b\left(b^{P+Q}\right)

    . . . . . . . . . . . . . . \log_b(xy) \:=\:(P+Q)\underbrace{\log_bb}_{\text{This is 1}}
    . . . . . . . . . . . . . . \log_b(xy) \:=\:P+Q


    Replace [1] and [2]: . \log_b(xy) \;=\;\log_b(x) + \log_b(y)

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  4. #4
    Member integral's Avatar
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    Thank you

    This is really great fun, I ended up writing proofs for all the basic logarithmic properties.
    ( I like your method undefined.)
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