# Math Help - My first proofs

1. ## 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
$
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?

2. Originally Posted by integral
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
$
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, $log_b(x^y)=ylog_bx$.

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

4. Thank you

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