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11-lg-4-lg-4-lg-2-16-lg-2-lg-2-3-12-5lg-3-3-lg-4-1-2-1-lg-2-8-lg-3-27-lg-2-1-2-




Question Number 58896 by cesar.marval.larez@gmail.com last updated on 03/May/19
11)  lg_4 lg_4 lg_2 16−lg_2 lg_2 (√3)  12. (5lg_3 3−lg_4 1)^2 +(((1/(lg_2 8))×lg_3 27)/(lg_(√2) (1/2)))
$$\left.\mathrm{11}\right)\:\:\mathrm{lg}_{\mathrm{4}} \mathrm{lg}_{\mathrm{4}} \mathrm{lg}_{\mathrm{2}} \mathrm{16}−\mathrm{lg}_{\mathrm{2}} \mathrm{lg}_{\mathrm{2}} \sqrt{\mathrm{3}} \\ $$$$\mathrm{12}.\:\left(\mathrm{5lg}_{\mathrm{3}} \mathrm{3}−\mathrm{lg}_{\mathrm{4}} \mathrm{1}\right)^{\mathrm{2}} +\frac{\frac{\mathrm{1}}{\mathrm{lg}_{\mathrm{2}} \mathrm{8}}×\mathrm{lg}_{\mathrm{3}} \mathrm{27}}{\mathrm{lg}_{\sqrt{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}}} \\ $$
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
$${lg}_{\mathrm{4}} {lg}_{\mathrm{4}} {lg}_{\mathrm{2}} \mathrm{16} \\ $$$${lg}_{\mathrm{4}} {lg}_{\mathrm{4}} {lg}_{\mathrm{2}} \mathrm{2}^{\mathrm{4}} \\ $$$${lg}_{\mathrm{4}} {lg}_{\mathrm{4}} \mathrm{4}\:\:\:\:\left[{lg}_{\mathrm{2}} \mathrm{2}=\mathrm{1}\right] \\ $$$${lg}_{\mathrm{4}} \mathrm{1} \\ $$$$=\mathrm{0} \\ $$$${lg}_{\mathrm{2}} {lg}_{\mathrm{2}} \mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${lg}_{\mathrm{2}} \mathrm{2}^{−\mathrm{1}} {lg}_{\mathrm{2}} \mathrm{3} \\ $$$$=\left(−\mathrm{1}\right)×{lg}_{\mathrm{2}} \mathrm{3} \\ $$$${ans}\:=\mathrm{0}−\left(−{lg}_{\mathrm{2}} \mathrm{3}\right)={lg}_{\mathrm{2}} \mathrm{3} \\ $$
Answered by tanmay last updated on 01/May/19
(5lg_3 3−lg_4 1)^2   =(5×1−0)^2 =25  (((1/(lg_2 8))×lg_3 27)/(lg_((√2) ) (1/2)))  =(((1/(lg_2 2^3 ))×lg_3 3^3 )/(lg_((√2) ) ((√2) )^(−2) ))  =((3/3)/(−2))=((−1)/2)  ans is 25−(1/2)  =((49)/2)
$$\left(\mathrm{5}{lg}_{\mathrm{3}} \mathrm{3}−{lg}_{\mathrm{4}} \mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{5}×\mathrm{1}−\mathrm{0}\right)^{\mathrm{2}} =\mathrm{25} \\ $$$$\frac{\frac{\mathrm{1}}{{lg}_{\mathrm{2}} \mathrm{8}}×{lg}_{\mathrm{3}} \mathrm{27}}{{lg}_{\sqrt{\mathrm{2}}\:} \frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\frac{\mathrm{1}}{{lg}_{\mathrm{2}} \mathrm{2}^{\mathrm{3}} }×{lg}_{\mathrm{3}} \mathrm{3}^{\mathrm{3}} }{{lg}_{\sqrt{\mathrm{2}}\:} \left(\sqrt{\mathrm{2}}\:\right)^{−\mathrm{2}} } \\ $$$$=\frac{\frac{\mathrm{3}}{\mathrm{3}}}{−\mathrm{2}}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${ans}\:{is}\:\mathrm{25}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{49}}{\mathrm{2}} \\ $$$$ \\ $$

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