Question

Given loga Xlogb X+logbXlogcX+logcXlogaX=logaXlogbXlogcX, prove that X=abc

Answer 1

use loga X= (log x)/( log a)

Answer 2

I'm going to rewrite this a sec for you, it's really hard to read.

Answer 3

(log x)/( log a) (log x)/( log b) +(log x)/( log b) (log x)/( log c ) + (log x)/( log a) (log x)/( log c) = (log x)/( log a) (log x)/( log b) (log x)/( log c)

Answer 4

Hopefully I didn't make any typos :O\[\large \log_a X \cdot \log_b X+\log_b X \cdot \log_c X+\log_c X \cdot \log_a X \quad =\]\[\large \log_a X \cdot \log_b X \cdot \log_c X\]

Answer 5

or
((logx)^2)(1/((loga)(logb) + 1/((logc)(logb) +1/((loga)(logc)) ) =((log x)^3 ) / ((loga)(logb)(logc))

((loga)(logb)(logc)) *(1/((loga)(logb) + 1/((logc)(logb) +1/((loga)(logc)) ) =(log x)
or (log c +log a+ logb)= logx
or log abc =log x
or x=abc

Answer 6

incase of typos error for braces you may check

Answer 7

Thanks,90 degree bow to u all. :-)