• question_answer Which of the following is correct? A)  If${{a}^{2}}+4{{b}^{2}}=12ab,$then  $\log (a+2b)=\frac{1}{2}(\log a+\log b)$B)  If$\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b},$then${{x}^{a}}.{{y}^{b}}.{{z}^{c}}=abc$C) $\frac{1}{{{\log }_{xy}}xyz}+\frac{1}{{{\log }_{yz}}xyz}+\frac{1}{{{\log }_{zx}}xyz}=2$D)  All are correct

[a]$\log (a+2b)=\frac{1}{2}\log {{(a+2b)}^{2}}$ $=\frac{1}{2}\log ({{a}^{2}}+4{{b}^{2}}+4ab)$ $=\frac{1}{2}\log (12ab+4ab)$ $=\frac{1}{2}\log ({{2}^{4}}.ab)$ $=\frac{1}{2}(4\log 2+\log a+\log b)$ [b] Let$\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k$ $\Rightarrow$            $\log x=k(b-c),\,\,\log y=k(c-a)$             $\log z=k(a-b)$ $\therefore$    ${{x}^{a}}.{{y}^{b}}.{{z}^{c}}={{p}^{k[a(b-c)+b(c-a)+c(a-b)]}}$ $={{p}^{k(0)}}=1$where $p$ is any arbitrary base of the log. [c] Given expression $={{\log }_{xyz}}xy+{{\log }_{xyz}}yz+{{\log }_{xyz}}zx$ $={{\log }_{xyz}}(xy.yz.zx)={{\log }_{xyz}}({{x}^{2}}.{{y}^{2}}.{{z}^{2}})$ $=2{{\log }_{xyz}}(xyz)=2\times 1=2$.