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Question Number 59401 by Pranay last updated on 09/May/19

5log_(4(√2)) (3−(√6) ) −6log_8 ((√3)−(√2))

$$\mathrm{5}{log}_{\mathrm{4}\sqrt{\mathrm{2}}} \left(\mathrm{3}−\sqrt{\mathrm{6}}\:\right)\:−\mathrm{6}{log}_{\mathrm{8}} \left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right) \\ $$

Answered by prakash jain last updated on 10/May/19

4(√2)=((√2))^5 8=((√2))^6 5log_(((√2))^5 ) (3−(√6))−6log_(((√2))^6 ) ((√3)−(√2)) Using log_a^n b=(1/n)log_a b 5(1/5)log_(√2) (3−(√6))−6(1/6)log_(√2) ((√3)−(√2)) =log_(√2) ((3−(√6))/((√3)−(√2))) =log_(√2) (((3−(√6))/((√3)−(√2)))×(((√3)+(√2))/((√3)+(√2)))) =log_(√2) (3(√3)−(√(18))+3(√2)−(√(12))) =log_(√2) (3(√3)−3(√2)+3(√2)−2(√3)) =log_(√2) (√3) =(1/2)log_(√2) 3 =(1/2)×(1/(1/2))log_2 3 =log_2 3

$$\mathrm{4}\sqrt{\mathrm{2}}=\left(\sqrt{\mathrm{2}}\right)^{\mathrm{5}} \\ $$$$\mathrm{8}=\left(\sqrt{\mathrm{2}}\right)^{\mathrm{6}} \\ $$$$\mathrm{5log}_{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{5}} } \left(\mathrm{3}−\sqrt{\mathrm{6}}\right)−\mathrm{6log}_{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{6}} } \left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right) \\ $$$$\:\:\:\mathrm{Using}\:\mathrm{log}_{{a}^{{n}} } {b}=\frac{\mathrm{1}}{{n}}\mathrm{log}_{{a}} {b} \\ $$$$\mathrm{5}\frac{\mathrm{1}}{\mathrm{5}}\mathrm{log}_{\sqrt{\mathrm{2}}} \left(\mathrm{3}−\sqrt{\mathrm{6}}\right)−\mathrm{6}\frac{\mathrm{1}}{\mathrm{6}}\mathrm{log}_{\sqrt{\mathrm{2}}} \left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right) \\ $$$$=\mathrm{log}_{\sqrt{\mathrm{2}}} \frac{\mathrm{3}−\sqrt{\mathrm{6}}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}} \\ $$$$=\mathrm{log}_{\sqrt{\mathrm{2}}} \left(\frac{\mathrm{3}−\sqrt{\mathrm{6}}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}}×\frac{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}}{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}}\right) \\ $$$$=\mathrm{log}_{\sqrt{\mathrm{2}}} \left(\mathrm{3}\sqrt{\mathrm{3}}−\sqrt{\mathrm{18}}+\mathrm{3}\sqrt{\mathrm{2}}−\sqrt{\mathrm{12}}\right) \\ $$$$=\mathrm{log}_{\sqrt{\mathrm{2}}} \left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}}\right) \\ $$$$=\mathrm{log}_{\sqrt{\mathrm{2}}} \sqrt{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{\sqrt{\mathrm{2}}} \mathrm{3} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{1}/\mathrm{2}}\mathrm{log}_{\mathrm{2}} \mathrm{3} \\ $$$$=\mathrm{log}_{\mathrm{2}} \mathrm{3} \\ $$

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