# Logarithms with a twist

MENTOR note: THread moved from General Math

I just can get it

This problem has been driving me crazy for a week now

$$If\quad a,b,c\quad \neq \quad 1\\ Also\quad (log_{ b }{ a\cdot }log_{ c }{ a }\quad +\quad log_{ a }{ a })\quad +\quad (log_{ a }{ b\cdot }log_{ c }{ b }\quad +\quad log_{ b }{ b })\quad +\quad (log_{ a }{ c\cdot }log_{ b }{ c }\quad +\quad log_{ c }{ c })\quad =\quad 0\\ Prove\quad that\quad \\ abc\quad =\quad 1\\ Additionally\\ If\quad a,b,c\quad =\quad 1\quad Prove\quad a=b=c$$

My attempt

So what I did was

call log_a b = p ,log_b = q,log_c a=q

I plugged this in and made an interesting discovery

pqr = 1

So i used this in the equation above and tried to get to the proof but didn't work

Where am I faltering
?

jedishrfu
Mentor
On the first proof, did you notice how to simplify the ##log_{a} ( a )##?

Of course i wrote it as 1 and proceeded

jedishrfu
Mentor
Did you try to convert the factors to a common log base base like log_b(a) + ln(a)/ln(b)?

Yes i did try that

jedishrfu
Mentor
and will you be showing your work? We can't help you here unless you show some work.

Well its alright i got it on my own

jedishrfu
jedishrfu
Mentor
so what was the trick you needed?

I needed to remember that If a3 + b3 + c3 = 3abc

a+b+c=0 or a=b=c

jedishrfu
Mentor
I needed to remember that If a3 + b3 + c3 = 3abc

a+b+c=0 or a=b=c

Where did that come from? It wasn't part of the original problem you posted.

haruspex