1. ## log proof

Prove that
if $\displaystyle \log_{x}y = am^{2}$ , for all a,b,m, bam = 1, then $\displaystyle \log_{y}x = \frac{b}{m}$

2. Originally Posted by hello
Prove that
if $\displaystyle \log_{x}y = am^{2}$ , for all a,b,m, bam = 1, then $\displaystyle \log_{y}x = \frac{b}{m}$

note that $\displaystyle am^2= am\cdot m = \frac{1}{b}\cdot m =\frac{m}{b}$

rewriting $\displaystyle \log_{x}y=am^2 \iff x^{am^2}=y$

then using the above sub we get

$\displaystyle x^{\frac{m}{b}}=y \iff x=y^{\frac{b}{m}}$

rewriting in log form we get

$\displaystyle \log_{y}(x)=\frac{b}{m}$

3. Hello, hello!

Are we allowed the "reciprocal theorem"? .$\displaystyle \log_b(a) \:=\:\frac{1}{\log_a(b)}$

Given: .$\displaystyle \log_{x}y = am^2,\:\text{ and for all } a,b,m\!:\;abm = 1$

Prove: .$\displaystyle \log_{y}x = \frac{b}{m}$

We have: .$\displaystyle abm = 1\quad\Rightarrow\quad a = \frac{1}{bm}$

Then: .$\displaystyle \log_xy \:=\:am^2\:=\:\left(\frac{1}{bm}\right)m^2 \quad\Rightarrow\quad \log_xy \:=\:\frac{m}{b}$

The theorem says: .$\displaystyle \log_xy \:=\:\frac{1}{\log_yx} \:=\:\frac{m}{b} \quad\Rightarrow\quad \log_yx \:=\:\frac{b}{m}$