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Thread: 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:

    $\displaystyle log_bxy=log_bx+lob_by\,\,\,\because$
    if
    $\displaystyle log_bx=n$
    $\displaystyle log_by=z$
    then
    $\displaystyle b^{z+n}=xy$
    and
    $\displaystyle z+n=log_bxy$
    so
    $\displaystyle
    b^{log_bxy}=xy$
    $\displaystyle xy=xy$


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



    Prove:

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

    Are theses proofs? Or not?
    Last edited by integral; Apr 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:

    $\displaystyle log_bxy=log_bx+lob_by\,\,\,\because$
    if
    $\displaystyle log_bx=n$
    $\displaystyle log_by=z$
    then
    $\displaystyle b^{z+n}=xy$
    and
    $\displaystyle z+n=log_bxy$
    so
    $\displaystyle
    b^{log_bxy}=xy$
    $\displaystyle xy=xy$


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



    Prove:

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

    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 $\displaystyle log_bxy=log_bx+log_by$ and not $\displaystyle xy = xy$!

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

    Here's how I would change your steps:

    Prove that $\displaystyle log_bxy=log_bx+log_by$.

    Proof:

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

    Then $\displaystyle b^{n+z}=b^nb^z=xy$

    so $\displaystyle n+z=log_bxy$.

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

    (Edited for clarity)

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

    Prove that $\displaystyle log_b(x^y)=ylog_bx$.

    Proof:

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

    Then by definition, $\displaystyle z=log_bx$.

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

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

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

    Taking the log base b of both sides, $\displaystyle log_b(x^y)=ylog_bx$.
    Last edited by undefined; Apr 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: .$\displaystyle \log_b(xy) \:=\: \log_b(x) + \log_b(y)$


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


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


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

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


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

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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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