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

Sandra Allison 2022-01-07 Answered
Given \(\displaystyle{\frac{{{\log{{x}}}}}{{{b}-{c}}}}={\frac{{{\log{{y}}}}}{{{c}-{a}}}}={\frac{{{\log{{z}}}}}{{{a}-{b}}}}\) show that \(\displaystyle{x}^{{{b}+{c}-{a}}}\cdot{y}^{{{c}+{a}-{b}}}\cdot{z}^{{{a}+{b}-{c}}}={1}\)

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Expert Answer

maul124uk
Answered 2022-01-08 Author has 737 answers
We have
\(\displaystyle{\frac{{{\log{{x}}}}}{{{b}-{c}}}}={\frac{{{\log{{y}}}}}{{{c}-{a}}}}={\frac{{{\log{{z}}}}}{{{a}-{b}}}}={t}\)
This gives us
\(\displaystyle{x}={e}^{{{t}{\left({b}-{c}\right)}}},{y}={e}^{{{t}{\left({c}-{a}\right)}}}\ \text{ and }\ {z}={e}^{{{t}{\left({a}-{b}\right)}}}\)
Hence,
\(\displaystyle{x}^{{{b}+{c}-{a}}}\cdot{y}^{{{c}+{a}-{b}}}\cdot{z}^{{{a}+{b}+{c}}}={e}^{{{\left({\left({b}-{c}\right)}{\left({b}+{c}-{a}\right)}+{\left({c}-{a}\right)}{\left({c}+{a}-{b}\right)}+{\left({a}-{b}\right)}{\left({a}+{b}-{c}\right)}\right)}}}\)
\(\displaystyle={e}^{{{t}{\left({b}^{{2}}-{c}^{{2}}-{a}{b}+{a}{c}+{c}^{{2}}-{a}^{{2}}-{b}{c}+{b}{a}+{a}^{{2}}-{b}^{{2}}-{a}{c}+{b}{c}\right)}}}={e}^{{0}}={1}\)
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Karen Robbins
Answered 2022-01-09 Author has 4170 answers
If you want to use your equations, here is a method.
Multiplying the equations together, we obtain:
\(\displaystyle{x}^{{{c}-{a}}}{y}^{{{a}-{b}}}{z}^{{{b}-{c}}}={y}^{{{b}-{c}}}{z}^{{{c}-{a}}}{x}^{{{a}-{b}}}\)
which gives after reordering:
Therefore it suffices to show that \(\displaystyle{x}^{{a}}{y}^{{b}}{z}^{{c}}={1}\)
Your first and third equations give \(\displaystyle{y}={x}^{{{\frac{{{c}-{a}}}{{{b}-{c}}}}}},{z}={x}^{{{\frac{{{a}-{b}}}{{{b}-{c}}}}}}\) This gives us:
\(\displaystyle{x}^{{a}}{y}^{{b}}{z}^{{c}}={x}^{{a}}{x}^{{{\frac{{{c}-{a}}}{{{b}-{c}}}}\times{b}}}{x}^{{{\frac{{{a}-{b}}}{{{b}-{c}}}}\times{c}}}={x}^{{{a}+{\frac{{{b}{c}-{b}{a}+{c}{a}-{b}{c}}}{{{b}-{c}}}}}}={x}^{{{a}-{a}}}={x}^{{0}}={1}\)
QED
0
Vasquez
Answered 2022-01-11 Author has 8850 answers

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

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