Question Number 58896 by cesar.marval.larez@gmail.com last updated on 03/May/19
$$\left.\mathrm{11}\right)\:\:\mathrm{lg}_{\mathrm{4}} \mathrm{lg}_{\mathrm{4}} \mathrm{lg}_{\mathrm{2}} \mathrm{16}−\mathrm{lg}_{\mathrm{2}} \mathrm{lg}_{\mathrm{2}} \sqrt{\mathrm{3}} \\ $$$$\mathrm{12}.\:\left(\mathrm{5lg}_{\mathrm{3}} \mathrm{3}−\mathrm{lg}_{\mathrm{4}} \mathrm{1}\right)^{\mathrm{2}} +\frac{\frac{\mathrm{1}}{\mathrm{lg}_{\mathrm{2}} \mathrm{8}}×\mathrm{lg}_{\mathrm{3}} \mathrm{27}}{\mathrm{lg}_{\sqrt{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}}} \\ $$
Answered by tanmay last updated on 01/May/19
![lg_4 lg_4 lg_2 16 lg_4 lg_4 lg_2 2^4 lg_4 lg_4 4 [lg_2 2=1] lg_4 1 =0 lg_2 lg_2 3^(1/2) lg_2 2^(−1) lg_2 3 =(−1)×lg_2 3 ans =0−(−lg_2 3)=lg_2 3](https://www.tinkutara.com/question/Q58906.png)
$${lg}_{\mathrm{4}} {lg}_{\mathrm{4}} {lg}_{\mathrm{2}} \mathrm{16} \\ $$$${lg}_{\mathrm{4}} {lg}_{\mathrm{4}} {lg}_{\mathrm{2}} \mathrm{2}^{\mathrm{4}} \\ $$$${lg}_{\mathrm{4}} {lg}_{\mathrm{4}} \mathrm{4}\:\:\:\:\left[{lg}_{\mathrm{2}} \mathrm{2}=\mathrm{1}\right] \\ $$$${lg}_{\mathrm{4}} \mathrm{1} \\ $$$$=\mathrm{0} \\ $$$${lg}_{\mathrm{2}} {lg}_{\mathrm{2}} \mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${lg}_{\mathrm{2}} \mathrm{2}^{−\mathrm{1}} {lg}_{\mathrm{2}} \mathrm{3} \\ $$$$=\left(−\mathrm{1}\right)×{lg}_{\mathrm{2}} \mathrm{3} \\ $$$${ans}\:=\mathrm{0}−\left(−{lg}_{\mathrm{2}} \mathrm{3}\right)={lg}_{\mathrm{2}} \mathrm{3} \\ $$
Answered by tanmay last updated on 01/May/19
$$\left(\mathrm{5}{lg}_{\mathrm{3}} \mathrm{3}−{lg}_{\mathrm{4}} \mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{5}×\mathrm{1}−\mathrm{0}\right)^{\mathrm{2}} =\mathrm{25} \\ $$$$\frac{\frac{\mathrm{1}}{{lg}_{\mathrm{2}} \mathrm{8}}×{lg}_{\mathrm{3}} \mathrm{27}}{{lg}_{\sqrt{\mathrm{2}}\:} \frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\frac{\mathrm{1}}{{lg}_{\mathrm{2}} \mathrm{2}^{\mathrm{3}} }×{lg}_{\mathrm{3}} \mathrm{3}^{\mathrm{3}} }{{lg}_{\sqrt{\mathrm{2}}\:} \left(\sqrt{\mathrm{2}}\:\right)^{−\mathrm{2}} } \\ $$$$=\frac{\frac{\mathrm{3}}{\mathrm{3}}}{−\mathrm{2}}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${ans}\:{is}\:\mathrm{25}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{49}}{\mathrm{2}} \\ $$$$ \\ $$