If $(y/z)^a * (z/x)^b*(x/y)^c = 1$ Then prove $(y/z)^(1/(b-c))=(z/x)^(1/(c-a)) = (x/y)^(1/(a-b))$ ?

Question

5695 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-08T11:20:51

Please see below.

As $(y/z)^a*(z/x)^b*(x/y)^c=1$

$x^(c-b)y^(a-c)z^(b-a)=1$ .........(A)

Let $(y/z)^(1/(b-c))=k$

then $y=k^(b-c)z$ and putting this in (A), we get

$x^(c-b)k^((b-c)(a-c))z^(a-c)z^(b-a)=1$

or $x^(c-b)k^((b-c)(a-c))z^(b-c)=1$

or $(z/x)^(b-c)=k^((b-c)(c-a))$ (-note change from $a-c$ to $c-a$ )

or $z/x=k^(c-a)$

or $(z/x)^(1/(c-a))=k$

Similarly $(x/y)^(1/(a-b))=k$

Hence $(y/z)^(1/(b-c))=(z/x)^(1/(c-a))=(x/y)^(1/(a-b))$

If (y/z)^a * (z/x)^b*(x/y)^c = 1 (y/z)^a * (z/x)^b*(x/y)^c = 1 Then prove (y/z)^(1/(b-c))=(z/x)^(1/(c-a)) = (x/y)^(1/(a-b))(y/z)^(1/(b-c))=(z/x)^(1/(c-a)) = (x/y)^(1/(a-b))?