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Question Number 24069 by Joel577 last updated on 12/Nov/17

$$\mathrm{Simplify}$$$$\frac{({\log }_2\sqrt{5}.{\log }_{25}20)+{\log }_4\sqrt{50}}{{\log }_{4}70-{\log }_{15}49}$$

Answered by $@ty@m last updated on 12/Nov/17

$$=\frac{\frac{\log \sqrt{5}}{\log 2}\times \frac{\log 20}{\log 25}+\frac{\log \sqrt{50}}{\log 4}}{\frac{\log 70}{\log 4}-\frac{\log 49}{\log 15}}$$$$=\frac{\frac{\log 5}{2\log 2}\times \frac{(\log 2+\log 10)}{2\log 5}+\frac{(\log 5+\log 10)}{4\log 2}}{\frac{(\log 7+\log 10)}{2\log 2}-\frac{2\log 7}{\log 15}}$$$$=\frac{\frac{(\log 2+1)}{4\log 2}+\frac{(\log 5+1)}{4\log 2}}{\frac{\log 15(\log 7+1)-4\log 7.\log 2}{2\log 2.\log 15}}$$$$=\frac{\frac{(\log 2+\log 5+1)}{4\log 2}}{\frac{\log 15(\log 7+1)-4\log 7.\log 2}{2\log 2.\log 15}}$$$$=\frac{\frac{2}{4\log 2}}{\frac{\log 15(\log 7+1)-4\log 7.\log 2}{2\log 2.\log 15}}$$$$=\frac{\frac{1}{2\log 2}}{\frac{\log 15(\log 7+1)-4\log 7.\log 2}{2\log 2.\log 15}}$$$$=\frac{\log 15}{\log 15(\log 7+1)-4\log 7.\log 2}$$$$$$

Commented by math solver last updated on 12/Nov/17

$$\mathrm{does}\mathrm{this}\mathrm{means}\mathrm{simplification}\{\mathrm{just}$$$$\mathrm{asking}\}\mathrm{coz}$$$$\mathrm{as}\mathrm{i}\mathrm{was}\mathrm{trying}\mathrm{i}\mathrm{could}\mathrm{not}\mathrm{get}\mathrm{a}\mathrm{single}$$$$\mathrm{answer}\mathrm{so}\mathrm{thought}\mathrm{i}\mathrm{would}\mathrm{be}\mathrm{wrong}$$$$\mathrm{somewhere}.$$

Commented by $@ty@m last updated on 12/Nov/17

$$Nomoresimplificationpossible.$$$$Atthemostyoucanfinditsnumerical$$$$valueusinglogtableorsci.caculator$$

Commented by $@ty@m last updated on 12/Nov/17

Commented by Joel577 last updated on 13/Nov/17

$$Yes,Ialsothoughttherewassomething$$$$wrongwiththisquestion$$

Commented by Joel577 last updated on 13/Nov/17

$$Btw,thankyouverymuch$$

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