If \(\cfrac{log\,a}{x+y-2z}=\cfrac{log\,b}{y+z-2x}=\cfrac{log\,c}{z+x

1.	If \(\cfrac{log\,a}{x+y-2z}=\cfrac{log\,b}{y+z-2x}=\cfrac{log\,c}{z+x-2y}\),log a/x+y-2z = log b/ y +z -2x = log c/z+x-2y, show that abc = 1.
Answer» Let \(\cfrac{log\,a}{x+y-2z}=\cfrac{log\,b}{y+z-2x}=\cfrac{log\,c}{z+x-2y}\) = k \(\therefore\) log a = k (x + y - 2z), log b = k (y + z - 2x), log c = k (z + x - 2y) We have to prove that abc = 1 i.e., to prove that log (abc) = log 1 i.e. , to prove that log a + log b + log c = 0 L.H.S.= log a + log b + log c = k (x + y -2z) + k (y + z -2x) + k (z + x -2y) = k (x + y -2z + y + z -2x + z + x -2y) = 0 = R.H.S.

If \(\cfrac{log\,a}{x+y-2z}=\cfrac{log\,b}{y+z-2x}=\cfrac{log\,c}{z+x-2y}\),log a/x+y-2z = log b/ y +z -2x = log c/z+x-2y, show that abc = 1.

Answer»

Let \(\cfrac{log\,a}{x+y-2z}=\cfrac{log\,b}{y+z-2x}=\cfrac{log\,c}{z+x-2y}\) = k

\(\therefore\) log a = k (x + y - 2z), log b = k (y + z - 2x),

log c = k (z + x - 2y)

We have to prove that abc = 1

i.e., to prove that log (abc) = log 1

i.e. , to prove that log a + log b + log c = 0

L.H.S.= log a + log b + log c

= k (x + y -2z) + k (y + z -2x) + k (z + x -2y)

= k (x + y -2z + y + z -2x + z + x -2y)

= 0

= R.H.S.