Why is there a twist in these logarithms?

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Homework Help Overview

The discussion revolves around a logarithmic equation involving three variables \(a\), \(b\), and \(c\) that are not equal to 1. The original poster is attempting to prove that the product \(abc = 1\) based on the given logarithmic expressions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to simplify the logarithmic expressions by substituting variables and exploring relationships between them. Participants question the simplification of certain logarithmic terms and suggest converting to a common log base. There is also a focus on the implications of specific algebraic identities related to sums and products of the variables.

Discussion Status

The discussion includes various attempts to manipulate the logarithmic expressions, with some participants providing suggestions for simplification. The original poster claims to have found a solution independently, but there is ongoing inquiry about the reasoning behind certain algebraic identities mentioned.

Contextual Notes

Participants note the importance of specific conditions under which the logarithmic identities hold, as well as the implications of assuming \(a\), \(b\), and \(c\) equal to 1. There is also a mention of a potential oversight regarding the algebraic identity involving cubes.

sidt36
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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=cMy 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
?
 
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On the first proof, did you notice how to simplify the ##log_{a} ( a )##?
 
Of course i wrote it as 1 and proceeded
 
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
 
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:smile:
 
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Likes   Reactions: jedishrfu
I needed to remember that If a3 + b3 + c3 = 3abc

a+b+c=0 or a=b=c
 
  • #10
sidt36 said:
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.
 
  • #11
sidt36 said:
I needed to remember that If a3 + b3 + c3 = 3abc
Shouldn't there be a minus sign in there?
 

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