Question Number 69457 by mhmd last updated on 23/Sep/19
$${find}\:{value}\:{log}\mathrm{40}/\mathrm{9}\:\:+\mathrm{4}{log}\mathrm{5}\:\:+\mathrm{2}{log}\mathrm{6}\:\:? \\ $$
Answered by MJS last updated on 23/Sep/19
$$\mathrm{log}\:\frac{\mathrm{40}}{\mathrm{9}}\:+\mathrm{4log}\:\mathrm{5}\:+\mathrm{2log}\:\mathrm{6}\:= \\ $$$$=\mathrm{log}\:\mathrm{40}\:−\mathrm{log}\:\mathrm{9}\:+\mathrm{4log}\:\mathrm{5}\:+\mathrm{2log}\:\mathrm{2}\:+\mathrm{2log}\:\mathrm{3}= \\ $$$$=\mathrm{3log}\:\mathrm{2}\:+\mathrm{log}\:\mathrm{5}\:−\mathrm{2log}\:\mathrm{3}\:+\mathrm{4log}\:\mathrm{5}\:+\mathrm{2log}\:\mathrm{2}\:+\mathrm{2log}\:\mathrm{3}= \\ $$$$=\mathrm{5log}\:\mathrm{2}\:+\mathrm{5log}\:\mathrm{5}\:=\mathrm{5log}\:\mathrm{10} \\ $$
Commented by kaivan.ahmadi last updated on 24/Sep/19
$$ \\ $$$$\mathrm{5}{log}\mathrm{10}=\mathrm{5}×\mathrm{1}=\mathrm{5} \\ $$
Commented by MJS last updated on 24/Sep/19
$$\mathrm{not}\:\mathrm{sure}\:\mathrm{about}\:\mathrm{the}\:\mathrm{usage}\:\mathrm{of}\:“\mathrm{log}'' \\ $$$$\mathrm{some}\:\mathrm{people}\:\mathrm{use}\:“\mathrm{ln}''\:\mathrm{for}\:\mathrm{log}_{\mathrm{e}} \:\mathrm{and}\:“\mathrm{log}''\:\mathrm{for} \\ $$$$\mathrm{log}_{\mathrm{10}} ;\:\mathrm{others}\:\mathrm{use}\:“\mathrm{log}''\:\mathrm{for}\:\mathrm{log}_{\mathrm{e}} \:\mathrm{and}\:“\mathrm{log}_{{b}} ''\:\mathrm{for} \\ $$$$\mathrm{all}\:\mathrm{other}\:\mathrm{bases}\:{b} \\ $$