a²+4b²=12ab then log(a+2b)=

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$$\begin{gathered} \mathrm{Given} \mathrm{that}: \\ {\mathrm{a}}^2+4{\mathrm{b}}^2=12\mathrm{ab} ;\left(\mathrm{ii}\right) \\ \mathrm{We} \mathrm{have}: \\ \log \left(\mathrm{a}+2\mathrm{b}\right) \\ =\frac{1}{2}\left\{2\log \left(\mathrm{a}+2\mathrm{b}\right)\right\} \\ =\frac{1}{2}\left\{\log {\left(\mathrm{a}+2\mathrm{b}\right)}^2\right\} \left[\mathrm{As} \mathrm{nlog}\left(\mathrm{m}\right)=\log \left({\mathrm{m}}^{\mathrm{n}}\right)\right] \\ =\frac{1}{2}\left\{\log \left({\mathrm{a}}^2+4{\mathrm{b}}^2+2.\mathrm{a}.\left(2\mathrm{b}\right)\right)\right\} \\ =\frac{1}{2}\left\{\log \left({\mathrm{a}}^2+4{\mathrm{b}}^2+4\mathrm{ab}\right)\right\} \\ =\frac{1}{2}\left\{\log \left(12\mathrm{ab}+4\mathrm{ab}\right)\right\} \left[\mathrm{Using} \left(\mathrm{i}\right)\right] \\ =\frac{1}{2}\log \left(16\mathrm{ab}\right) \\ =\frac{1}{2}\left\{\mathrm{loga}+\mathrm{logb}+\log 16\right\} \left[\mathrm{As} \log \left(\mathrm{a}.\mathrm{b}.\mathrm{c}.....\right)=\mathrm{loga}+\mathrm{logb}+\mathrm{logc}+...\right] \\ =\frac{1}{2}\left\{\mathrm{loga}+\mathrm{logb}+\log 2^4\right\} \\ =\frac{1}{2}\left\{\mathrm{loga}+\mathrm{logb}+4\log 2\right\} \\ \mathrm{Hence} \mathrm{option}\left(\mathrm{c}\right) \mathrm{is} \mathrm{correct}. \end{gathered}$$

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