Given:
$$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}$$
We have to show that :
$$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$$
I made three equations using cross multiplication :
$$1.~~x^{c-a}=y^{b-c}$$ $$2.~~y^{a-b}=z^{c-a}$$ $$3.~~z^{b-c}=x^{a-b}$$
How do I proceed hereafter? If I multiply the equations, one variable goes away from exponents.
Thank you.
On
Given: $$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}=\lambda$$ we have: $$ x = e^{\lambda(b-c)},\quad y=e^{\lambda(c-a)},\quad z=e^{\lambda(a-b)}, $$ hence: $$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = \exp\left(\lambda\cdot\sum_{cyc}\left(b^2-c^2-a(b-c)\right)\right)=\exp(0)=1.$$
On
$\dfrac{\log x}{b-c}$ is equal to $\log\left(x^{1/(b-c)}\right)$ regardless of what the base of the logarithm is. Hence we have $$ \log\left(x^{1/(b-c)}\right) = \log\left(y^{1/(c-a)}\right) = \log\left(z^{1/(a-b)}\right). $$ Since the logarithm function is one-to-one, this entails $$ x^{1/(b-c)} = y^{1/(c-a)} = z^{1/(a-b)}. $$ Raising both sides of $x^{1/(b-c)} = y^{1/(c-a)}$ to the power $(b-c)(c-a)$ yields $$ x^{c-a} = y^{b-c} $$ and the other two equalities are derived similarly.
On
Hint
If you have to show that $$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$$ taking logarithms of both sides means that you have to show that $$(b+c-a)\log(x)+(c+a-b)\log(y)+(a+b-c)\log(z)=0$$ Now, use what user17762 and Jack D'Aurizio answered.
On
The product of the three equalities you have given is:
$x^{c-a} y^{a-b} z^{b-c} = y^{b-c} z^{c-a} x^{a-b}$
Grouping similar variables together,
$(x^{c-a} x^{b-a}) (y^{a-b} y^{c-b}) (z^{b-c} z^{a-c}) = 1$
$x^{-2a+b+c} \times y^{a-2b+c} \times z^{a+b-2c} = 1$
On
If you want to use your equations, here is a method.
Multiplying the equations together, we obtain: $$x^{c-a}y^{a-b}z^{b-c}=y^{b-c}z^{c-a}x^{a-b}$$ which gives after reordering: $$x^{b+c-a}y^{c+a-b}z^{a+b-c}=x^a y^b z^c$$ Therefore it suffices to show that $x^a y^b z^c = 1$.
Your first and third equations give $y = x^{\frac {c-a}{b-c}}, z = x^{\frac{a-b}{b-c}}$. This gives us: $$x^a y^b z^c = x^a x^{\frac {c-a}{b-c}\times b} x^{\frac{a-b}{b-c}\times c} = x^{a + \frac{bc-ba+ca-bc}{b-c}} = x^{a-a} = x^0 = 1 $$
QED.
On
First let's assume that there are no indeterminations:
$x > 0 \land y > 0 \land z > 0 \land a \neq b \neq c$
And we have to prove that $k = 1$ in:
$x^{b+c−a}⋅y^{c+a−b}⋅z^{a+b−c} = k$
Using the asker equations from cross multiplication:
$x^{b+c−a}⋅x^{a-c}⋅y^{a}⋅z^{a}⋅x^{a-b} = x^a⋅y^a⋅z^a = (xyz)^a = k$
From this method, we can also obtain:
$(xyz)^b = k$
and
$(xyz)^c = k$
But then:
$k^{\frac{1}a} = k^{\frac{1}b} = k^{\frac{1}c} \implies k = 1$
What's more:
$xyz = 1$
We have $$\dfrac{\log(x)}{b-c} = \dfrac{\log(y)}{c-a} = \dfrac{\log(z)}{a-b} = t$$ This gives us $$x=e^{t(b-c)}, y = e^{t(c-a)} \text{ and }z = e^{t(a-b)}$$ Hence, \begin{align} x^{b+c-a}\cdot y^{c+a-b} \cdot z^{a+b-c} & = e^{t\left((b-c)(b+c-a) + (c-a)(c+a-b) + (a-b)(a+b-c)\right)}\\ & = e^{t(b^2-c^2-ab+ac + c^2 - a^2 -bc + ba + a^2 - b^2 - ac + bc)} = e^0 = 1 \end{align}