Without-using-a-calculator-evaluate-log-5-2-log-2-log-50-

Question Number 5579 by Rasheed Soomro last updated on 21/May/16

$$\mathrm{Without}\:\mathrm{using}\:\mathrm{a}\:\mathrm{calculator},\:\mathrm{evaluate} \\ $$$$\left(\mathrm{log}\:\mathrm{5}\right)^{\mathrm{2}} +\mathrm{log}\:\mathrm{2}\:\mathrm{log}\:\mathrm{50} \\ $$

Answered by FilupSmith last updated on 21/May/16

$$\mathrm{log}\:{x}\:=\:\mathrm{log}_{\mathrm{10}} {x} \\ $$$$ \\ $$$$\mathrm{log}\left(\mathrm{5}\right)^{\mathrm{2}} +\mathrm{log}\left(\mathrm{5}\right)\mathrm{log}\left(\mathrm{50}\right) \\ $$$$=\mathrm{log}\left(\mathrm{5}\right)\left(\mathrm{log}\left(\mathrm{5}\right)+\mathrm{log}\left(\mathrm{50}\right)\right) \\ $$$$=\mathrm{log}\left(\mathrm{5}\right)\mathrm{log}\left(\mathrm{250}\right) \\ $$$$\mathrm{log}\left(\mathrm{250}\right)=\mathrm{log}\left(\mathrm{100}×\mathrm{2}.\mathrm{5}\right) \\ $$$$=\mathrm{log}\left(\mathrm{100}\right)+\mathrm{log}\left(\mathrm{2}.\mathrm{5}\right) \\ $$$$=\mathrm{2}+\mathrm{log}\left(\mathrm{2}.\mathrm{5}\right) \\ $$$$\therefore\mathrm{log}\left(\mathrm{5}\right)\mathrm{log}\left(\mathrm{250}\right)=\mathrm{log}\left(\mathrm{5}\right)\left(\mathrm{2}+\mathrm{log}\left(\mathrm{2}.\mathrm{5}\right)\right) \\ $$$$=\mathrm{2log}\left(\mathrm{5}\right)+\mathrm{log}\left(\mathrm{5}\right)\mathrm{log}\left(\mathrm{2}.\mathrm{5}\right) \\ $$$${continue} \\ $$

Commented by nchejane last updated on 21/May/16

$${It}\:{is}\:{like}\:{it}\:{is}\:{not}\:{copied}\:{correctly} \\ $$$$\left({log}\mathrm{5}\right)^{\mathrm{2}} +{log}\mathrm{2}{log}\mathrm{50} \\ $$$$=\left({log}\mathrm{5}\right)^{\mathrm{2}} +{log}\mathrm{2}\left({log}\left(\mathrm{5}×\mathrm{10}\right)\right) \\ $$$$=\left({log}\mathrm{5}\right)^{\mathrm{2}} +{log}\mathrm{2}\left({log}\mathrm{5}+{log}\mathrm{10}\right) \\ $$$$=\left({log}\mathrm{5}\right)^{\mathrm{2}} +{log}\mathrm{2}{log}\mathrm{5}+{log}\mathrm{2} \\ $$$$=\left({log}\mathrm{5}\right)\left[{log}\mathrm{5}+{log}\mathrm{2}\right]+{log}\mathrm{2} \\ $$$$={log}\mathrm{5}{log}\mathrm{10}+{log}\mathrm{2} \\ $$$$={log}\mathrm{5}+{log}\mathrm{2} \\ $$$$={log}\mathrm{10} \\ $$$$=\mathrm{1} \\ $$

Commented by Rasheed Soomro last updated on 21/May/16

$$\mathbb{G}^{\overset{\mathrm{v}} {\mathcal{OO}}} \mathbb{D}! \\ $$

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