Sandra Allison

2022-01-07

Given $\frac{\mathrm{log}x}{b-c}=\frac{\mathrm{log}y}{c-a}=\frac{\mathrm{log}z}{a-b}$ show that ${x}^{b+c-a}\cdot {y}^{c+a-b}\cdot {z}^{a+b-c}=1$

maul124uk

We have
$\frac{\mathrm{log}x}{b-c}=\frac{\mathrm{log}y}{c-a}=\frac{\mathrm{log}z}{a-b}=t$
This gives us

Hence,
${x}^{b+c-a}\cdot {y}^{c+a-b}\cdot {z}^{a+b+c}={e}^{\left(\left(b-c\right)\left(b+c-a\right)+\left(c-a\right)\left(c+a-b\right)+\left(a-b\right)\left(a+b-c\right)\right)}$
$={e}^{t\left({b}^{2}-{c}^{2}-ab+ac+{c}^{2}-{a}^{2}-bc+ba+{a}^{2}-{b}^{2}-ac+bc\right)}={e}^{0}=1$

Karen Robbins

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:
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:
${x}^{a}{y}^{b}{z}^{c}={x}^{a}{x}^{\frac{c-a}{b-c}×b}{x}^{\frac{a-b}{b-c}×c}={x}^{a+\frac{bc-ba+ca-bc}{b-c}}={x}^{a-a}={x}^{0}=1$
QED

Vasquez

Given:
$\frac{\mathrm{log}x}{b-c}=\frac{\mathrm{log}y}{c-a}=\frac{\mathrm{log}z}{a-b}=\lambda$
we have:
$x={e}^{\lambda \left(b+c\right)},y={e}^{\lambda \left(c-a\right)},z={e}^{\lambda \left(a-b\right)}$
hence:
${x}^{b+c-a}\cdot {y}^{c+a-b}\cdot {z}^{a+b-c}=\mathrm{exp}\left(\lambda \cdot \sum _{cyc}\left({b}^{2}-{c}^{2}-a\left(b-c\right)\right)\right)$
$=\mathrm{exp}\left(0\right)=1$

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