Question Number 57607 by Gulay last updated on 08/Apr/19
Commented by Gulay last updated on 08/Apr/19
$$\mathrm{sir}\:\mathrm{plz}\:\mathrm{help}\:\mathrm{me} \\ $$
Commented by MJS last updated on 08/Apr/19
$$\mathrm{used}\:\mathrm{rules}: \\ $$$$\mathrm{log}_{{b}} \:{a}\:=\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:{b}} \\ $$$$\mathrm{log}_{{b}} \:{a}^{{r}} \:={r}\mathrm{log}_{{b}} \:{a} \\ $$$$\sqrt[{{n}}]{{a}}={a}^{\frac{\mathrm{1}}{{n}}} \\ $$$${o} \\ $$$$\mathrm{log}_{\sqrt{\mathrm{2}}} \:\mathrm{16}\:=\frac{\mathrm{ln}\:\mathrm{16}}{\mathrm{ln}\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{ln}\:\mathrm{16}\:=\mathrm{ln}\:\mathrm{2}^{\mathrm{4}} \:=\mathrm{4ln}\:\mathrm{2} \\ $$$$\mathrm{ln}\:\sqrt{\mathrm{2}}\:=\mathrm{ln}\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{16}}{\mathrm{ln}\:\sqrt{\mathrm{2}}}=\frac{\mathrm{8ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{8} \\ $$$$... \\ $$
Commented by maxmathsup by imad last updated on 08/Apr/19
$${log}_{\sqrt{\mathrm{2}}} \left(\mathrm{16}\right)\:=\frac{{ln}\left(\mathrm{16}\right)}{{ln}\left(\sqrt{\mathrm{2}}\right)}\:=\frac{{ln}\left(\mathrm{2}^{\mathrm{4}} \right)}{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)}\:=\frac{\mathrm{8}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{2}\right)}\:=\mathrm{8}\:\:{also} \\ $$$${log}_{\sqrt{\mathrm{3}}} \left(\mathrm{9}\right)\:=\frac{{ln}\left(\mathrm{9}\right)}{{ln}\left(\sqrt{\mathrm{3}}\right)}\:=\frac{{ln}\left(\mathrm{3}^{\mathrm{2}} \right)}{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{3}\right)}\:=\frac{\mathrm{4}{ln}\left(\mathrm{3}\right)}{{ln}\left(\mathrm{3}\right)}\:=\mathrm{4}\:\Rightarrow \\ $$$${log}_{\sqrt{\mathrm{2}}} \left(\mathrm{16}\right)−{log}_{\sqrt{\mathrm{3}}} \left(\mathrm{9}\right)\:=\mathrm{8}−\mathrm{4}\:=\mathrm{4}\:\:\:\:{we}\:{remember}\:{that}\:{log}_{{a}} \left({x}\right)=\frac{{ln}\left({x}\right)}{{ln}\left({a}\right)} \\ $$
Answered by $@ty@m last updated on 10/Apr/19
$$\mathrm{log}\:_{\sqrt{\mathrm{2}}} \mathrm{16}−\mathrm{log}\:_{\sqrt{\mathrm{3}}} \mathrm{9} \\ $$$$=\mathrm{log}\:_{\sqrt{\mathrm{2}}} \left(\sqrt{\mathrm{2}}\right)^{\mathrm{8}} −\mathrm{log}\:_{\sqrt{\mathrm{3}}} \left(\sqrt{\mathrm{3}}\right)^{\mathrm{4}} \\ $$$$=\mathrm{8}−\mathrm{4} \\ $$$$=\mathrm{4} \\ $$
Commented by Gulay last updated on 08/Apr/19
$$\mathrm{thanks}\:\mathrm{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 08/Apr/19
$${Good}\:{approach}!...{Minor}\:{error}. \\ $$$$\mathrm{log}\:_{\sqrt{\mathrm{2}}} \mathrm{16}−\mathrm{log}\:_{\sqrt{\mathrm{3}}} \mathrm{9} \\ $$$$=\mathrm{log}\:_{\sqrt{\mathrm{2}}} \left(\sqrt{\mathrm{2}}\right)^{\mathrm{8}} −\mathrm{log}\:_{\sqrt{\mathrm{3}}} \left(\sqrt{\mathrm{3}}\right)^{\mathrm{4}} \\ $$$$=\mathrm{8}−\mathrm{4}=\mathrm{4} \\ $$$$ \\ $$
Commented by $@ty@m last updated on 10/Apr/19
$${Thanks}\:{for}\:{correction}. \\ $$