Question

Solve this:
Q.10. Calculate : $4^{5{\log }_{4\sqrt{2}}\left(3-\sqrt{6}\right)-6{\log }_8\left(\sqrt{3}-\sqrt{2}\right)}$

Answer 1

Dear student
$$\begin{gathered} 4^{5{\log }_{4\sqrt{2}}\left(3-\sqrt{6}\right)-6{\log }_8\left(\sqrt{3}-\sqrt{2}\right)} \\ =4^{-6{\log }_8\left(\sqrt{3}-\sqrt{2}\right)}\times 4^{5{\log }_{4\sqrt{2}}\left(3-\sqrt{6}\right)}\left[\because {\mathrm{a}}^{\mathrm{b}+\mathrm{c}}={\mathrm{a}}^{\mathrm{b}}\times {\mathrm{a}}^{\mathrm{c}}\right] \\ \mathrm{Consider},4^{-6{\log }_8\left(\sqrt{3}-\sqrt{2}\right)} \\ ={\left(2^2\right)}^{-6{\log }_8\left(\sqrt{3}-\sqrt{2}\right)} \\ =2^{-6\times 2{\log }_8\left(\sqrt{3}-\sqrt{2}\right)}\left[\because {\left({\mathrm{a}}^{\mathrm{b}}\right)}^{\mathrm{c}}={\mathrm{a}}^{\mathrm{bc}}\right] \\ =2^{-6\times 2\times \frac{1}{3}{\log }_2\left(\sqrt{3}-\sqrt{2}\right)}\left[\because {\log }_{{\mathrm{a}}^{\mathrm{b}}}\left(\mathrm{x}\right)=\frac{1}{\mathrm{b}}{\log }_{\mathrm{a}}\left(\mathrm{x}\right)\right] \\ =2^{-4{\log }_2\left(\sqrt{3}-\sqrt{2}\right)} \\ ={\left(2^{{\log }_2\left(\sqrt{3}-\sqrt{2}\right)}\right)}^{-4} \\ ={\left(\sqrt{3}-\sqrt{2}\right)}^{-4}\left[\because {\mathrm{a}}^{{\log }_{\mathrm{a}}\left(\mathrm{b}\right)}=\mathrm{b}\right] \\ \mathrm{So},4^{-6{\log }_8\left(\sqrt{3}-\sqrt{2}\right)}\times 4^{5{\log }_{4\sqrt{2}}\left(3-\sqrt{6}\right)} \\ =4^{5{\log }_{4\sqrt{2}}\left(3-\sqrt{6}\right)}\times \frac{1}{{\left(\sqrt{3}-\sqrt{2}\right)}^4} \\ ={\left(4^5\right)}^{{\log }_{4\sqrt{2}}\left(3-\sqrt{6}\right)}\times \frac{1}{{\left(\sqrt{3}-\sqrt{2}\right)}^4} \\ ={1024}^{{\log }_{4\sqrt{2}}\left(3-\sqrt{6}\right)}\times \frac{1}{{\left(\sqrt{3}-\sqrt{2}\right)}^4} \\ =\frac{{1024}^{{\log }_{4\sqrt{2}}\left(3-\sqrt{6}\right)}}{{\left(\sqrt{3}-\sqrt{2}\right)}^4} \end{gathered}$$
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