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

Simplify (((log_2 (√5) . log_(25) 20) + log_4 (√(50)) )/(log_4 70 − log_(15) 49))

$$\mathrm{Simplify} \\ $$$$\frac{\left(\mathrm{log}_{\mathrm{2}} \:\sqrt{\mathrm{5}}\:.\:\mathrm{log}_{\mathrm{25}} \:\mathrm{20}\right)\:+\:\mathrm{log}_{\mathrm{4}} \:\sqrt{\mathrm{50}}\:\:}{\mathrm{log}_{\mathrm{4}} \:\mathrm{70}\:−\:\mathrm{log}_{\mathrm{15}} \:\mathrm{49}} \\ $$

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

=((((log (√5))/(log 2))×((log 20)/(log 25))+((log (√(50)))/(log 4)))/(((log 70)/(log 4))−((log 49)/(log 15)))) =((((log 5)/(2log 2))×(((log 2+log 10))/(2log 5))+(((log 5+log 10))/(4log 2)))/((((log 7+log 10))/(2log 2))−((2log 7)/(log15)))) =(((((log 2+1))/(4log 2))+(((log 5+1))/(4log 2)))/((log 15(log 7+1)−4log 7.log 2)/(2log 2.log 15))) =((((log 2+log 5+1))/(4log 2))/((log 15(log 7+1)−4log 7.log 2)/(2log 2.log 15))) =((2/(4log 2))/((log 15(log 7+1)−4log 7.log 2)/(2log 2.log 15))) =((1/(2log 2))/((log 15(log 7+1)−4log 7.log 2)/(2log 2.log 15))) =((log 15)/(log 15(log 7+1)−4log 7.log 2))

$$=\frac{\frac{\mathrm{log}\:\sqrt{\mathrm{5}}}{\mathrm{log}\:\mathrm{2}}×\frac{\mathrm{log}\:\mathrm{20}}{\mathrm{log}\:\mathrm{25}}+\frac{\mathrm{log}\:\sqrt{\mathrm{50}}}{\mathrm{log}\:\mathrm{4}}}{\frac{\mathrm{log}\:\mathrm{70}}{\mathrm{log}\:\mathrm{4}}−\frac{\mathrm{log}\:\mathrm{49}}{\mathrm{log}\:\mathrm{15}}} \\ $$$$=\frac{\frac{\mathrm{log}\:\mathrm{5}}{\mathrm{2log}\:\mathrm{2}}×\frac{\left(\mathrm{log}\:\mathrm{2}+\mathrm{log}\:\mathrm{10}\right)}{\mathrm{2log}\:\mathrm{5}}+\frac{\left(\mathrm{log}\:\mathrm{5}+\mathrm{log}\:\mathrm{10}\right)}{\mathrm{4log}\:\mathrm{2}}}{\frac{\left(\mathrm{log}\:\mathrm{7}+\mathrm{log}\:\mathrm{10}\right)}{\mathrm{2log}\:\mathrm{2}}−\frac{\mathrm{2log}\:\mathrm{7}}{\mathrm{log15}}} \\ $$$$=\frac{\frac{\left(\mathrm{log}\:\mathrm{2}+\mathrm{1}\right)}{\mathrm{4log}\:\mathrm{2}}+\frac{\left(\mathrm{log}\:\mathrm{5}+\mathrm{1}\right)}{\mathrm{4log}\:\mathrm{2}}}{\frac{\mathrm{log}\:\mathrm{15}\left(\mathrm{log}\:\mathrm{7}+\mathrm{1}\right)−\mathrm{4log}\:\mathrm{7}.\mathrm{log}\:\mathrm{2}}{\mathrm{2log}\:\mathrm{2}.\mathrm{log}\:\mathrm{15}}} \\ $$$$=\frac{\frac{\left(\mathrm{log}\:\mathrm{2}+\mathrm{log}\:\mathrm{5}+\mathrm{1}\right)}{\mathrm{4log}\:\mathrm{2}}}{\frac{\mathrm{log}\:\mathrm{15}\left(\mathrm{log}\:\mathrm{7}+\mathrm{1}\right)−\mathrm{4log}\:\mathrm{7}.\mathrm{log}\:\mathrm{2}}{\mathrm{2log}\:\mathrm{2}.\mathrm{log}\:\mathrm{15}}} \\ $$$$=\frac{\frac{\mathrm{2}}{\mathrm{4log}\:\mathrm{2}}}{\frac{\mathrm{log}\:\mathrm{15}\left(\mathrm{log}\:\mathrm{7}+\mathrm{1}\right)−\mathrm{4log}\:\mathrm{7}.\mathrm{log}\:\mathrm{2}}{\mathrm{2log}\:\mathrm{2}.\mathrm{log}\:\mathrm{15}}} \\ $$$$=\frac{\frac{\mathrm{1}}{\mathrm{2log}\:\mathrm{2}}}{\frac{\mathrm{log}\:\mathrm{15}\left(\mathrm{log}\:\mathrm{7}+\mathrm{1}\right)−\mathrm{4log}\:\mathrm{7}.\mathrm{log}\:\mathrm{2}}{\mathrm{2log}\:\mathrm{2}.\mathrm{log}\:\mathrm{15}}} \\ $$$$=\frac{\mathrm{log}\:\mathrm{15}}{\mathrm{log}\:\mathrm{15}\left(\mathrm{log}\:\mathrm{7}+\mathrm{1}\right)−\mathrm{4log}\:\mathrm{7}.\mathrm{log}\:\mathrm{2}} \\ $$$$ \\ $$

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

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

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

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

No more simplification possible. At the most you can find its numerical value using log table or sci.caculator

$${No}\:{more}\:{simplification}\:{possible}. \\ $$$${At}\:{the}\:{most}\:{you}\:{can}\:{find}\:{its}\:{numerical}\: \\ $$$${value}\:{using}\:{log}\:{table}\:{or}\:{sci}.{caculator} \\ $$

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

Commented by Joel577 last updated on 13/Nov/17

Yes, I also thought there was something wrong with this question

$${Yes},\:{I}\:{also}\:{thought}\:{there}\:{was}\:{something} \\ $$$${wrong}\:{with}\:{this}\:{question} \\ $$

Commented by Joel577 last updated on 13/Nov/17

Btw, thank you very much

$${Btw},\:{thank}\:{you}\:{very}\:{much} \\ $$

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