A) \[abc\]
B) \[xyz\]
C) \[a\]
D) \[\frac{1}{a}\]
Correct Answer: B
Solution :
Let \[\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=z\] \[\Rightarrow \] \[\log x=k(b-c)\] \[\Rightarrow \] \[x={{e}^{k(b-c)}}\] Similarly, \[y={{e}^{k(c-a)}},z={{e}^{k(a-b)}}\] \[\therefore \] \[{{x}^{a}}.{{y}^{b}}.{{z}^{c}}={{e}^{ak(b-c)}}.{{e}^{bk(c-a)}}.{{e}^{ck(a-b)}}\] \[={{e}^{k[a(b-c)+b(c-a)+c(a-b)]}}\] \[={{e}^{k[ab-ac+bc-ab+ca-bc]}}\] \[={{e}^{k[0]}}={{e}^{0}}\] \[\Rightarrow \] \[{{x}^{a}},{{y}^{b}}.{{z}^{c}}=1\] ?.(i) Now, \[log\text{ }x+log\text{ }y+log\text{ }z=k(b-c)\] \[+k(c-a)+k(a-b)\] \[\Rightarrow \] \[log\text{ }xyz=k[b-c+c-a+a-b]\] \[\Rightarrow \] \[log\,xyz=0\] \[\Rightarrow \] \[xyz={{e}^{0}}=1\] ?.(ii) From Eqs. (i) and (ii) \[{{x}^{a}},{{y}^{b}}.{{z}^{c}}=xyz\]
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