1. ## Simple Log Proofs

1) Prove that $\log_{b} a + \log_{c} b + \log_{a} c = \displaystyle{\frac{1}{\log_{a} b}} + \displaystyle{\frac{1}{\log_{b} c}} + \displaystyle{\frac{1}{\log_{c} a}}$

2) Prove that if $\log_{r} p = q$ and $\log_{q} r = p$, then $\log_{q} p = pq$

If you could point me in the right direction I should be able to take it from there.

Thanks again.

2. ## Re: Simple Log Proofs

log b (a) =x
b^x =a
loga((b^x)=loga (a)
x(log a(b)=1
x=1/loga(b)

3. ## Re: Simple Log Proofs

Originally Posted by bjhopper
log b (a) =x
b^x =a
loga((b^x)=loga (a)
x(log a(b)=1
x=1/loga(b)
@bjhopper, Why in the world do you not some effort to learn the code in LaTeX?

\$\log_b(a)\$ gives $\log_b(a)$.

4. ## Re: Simple Log Proofs

Originally Posted by bjhopper
log b (a) =x
b^x =a
loga((b^x)=loga (a)
x(log a(b)=1
x=1/loga(b)
Damn. I thought it would be this, but I was weary of adding in an 'x'. Thank you.

5. ## Re: Simple Log Proofs

Hello, Fratricide!

$\text{(2) Prove that if }\:\begin{Bmatrix}\log_{r} p &=& q & [1] \\ \log_{q} r &=& p & [2]\end{Bmatrix},\;\text{then: }\:\log_{q} p \:=\: p\!\cdot\!q$

$\text{From [1]: }\;\log_r(p) \,=\,q \quad\Rightarrow\quad p \:=\:r^q$

$\text{Take logs, base }q\!:\;\;\log_q(p) \:=\:\log_q(r^q) \quad\Rightarrow\quad \log_q(p) \:=\:q\!\cdot\!\underbrace{\log_q(r)}_{\text{This is }p}$
$\text{Therefore: }\;\log_q(p) \;=\;p\!\cdot\!q$