# My first proofs

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• Apr 5th 2010, 02:42 AM
integral
My first proofs
I had never heard of a proof before coming to this forum (Schools FTW (Lipssealed)).
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? (Bow)
• Apr 5th 2010, 03:51 AM
undefined
Quote:

Originally Posted by integral
I had never heard of a proof before coming to this forum (Schools FTW (Lipssealed)).
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? (Bow)

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, $\displaystyle log_b(x^y)=ylog_bx$.
• Apr 5th 2010, 06:15 AM
Soroban
Hello, integral!

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

Quote:

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]: .$\displaystyle \log_b(xy) \;=\;\log_b(x) + \log_b(y)$

• Apr 5th 2010, 01:12 PM
integral
Thank you (Giggle)

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