Question Number 215840 by golsendro last updated on 19/Jan/25
$$\:\:\:\mathrm{log}\:_{\mathrm{24}} \:\mathrm{3}=\:{a}\:\mathrm{and}\:\mathrm{log}\:_{\mathrm{24}} \:\mathrm{6}\:=\:\frac{{b}}{\mathrm{6}} \\ $$$$\:\:\:\mathrm{log}\:_{\sqrt{\mathrm{8}}} \:\left({b}−\mathrm{4}{a}\right)=\:? \\ $$
Commented by oubiji last updated on 19/Jan/25
$$\:\:\:\:{b}−\mathrm{4}{a}=\mathrm{log}_{\mathrm{24}} \mathrm{6}^{\mathrm{6}} −{log}_{\mathrm{24}} \mathrm{3}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={log}_{\mathrm{24}} \frac{\mathrm{6}^{\mathrm{6}} }{\mathrm{3}^{\mathrm{4}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={log}_{\mathrm{24}} \mathrm{24}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2} \\ $$$$\:\:\:{then}\:: \\ $$$$\:\:\:\:\:{log}_{\sqrt{\mathrm{8}}} \left({b}−\mathrm{4}{a}\right)={log}_{\sqrt{\mathrm{8}}} \mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{ln}\mathrm{2}}{{ln}\sqrt{\mathrm{8}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{ln}\mathrm{2}}{\frac{\mathrm{3}}{\mathrm{2}}{ln}\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 20/Jan/25
$${please}\:{next}\:{time}\:\:{post}\:{your}\:{solution}\:{as}\: \\ $$$$“{answer}''\:{not}\:{as}\:{a}\:“{comment}'' \\ $$
Answered by A5T last updated on 19/Jan/25
$$\mathrm{b}=\mathrm{log}_{\mathrm{24}} \mathrm{6}^{\mathrm{6}} \:\:;\:\:\:\:\mathrm{4a}=\mathrm{log}_{\mathrm{24}} \mathrm{3}^{\mathrm{4}} \\ $$$$\mathrm{b}−\mathrm{4a}=\mathrm{log}_{\mathrm{24}} \left(\frac{\mathrm{6}^{\mathrm{6}} }{\mathrm{3}^{\mathrm{4}} }\right)=\mathrm{log}_{\mathrm{24}} \left(\mathrm{24}^{\mathrm{2}} \right)=\mathrm{2} \\ $$$$\Rightarrow\mathrm{log}_{\sqrt{\mathrm{8}}} \left(\mathrm{b}−\mathrm{4a}\right)\Rightarrow\mathrm{log}_{\sqrt{\mathrm{8}}} \left(\mathrm{2}\right)=\mathrm{log}_{\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} } \left(\mathrm{2}\right)=\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Answered by MathematicalUser2357 last updated on 26/Jan/25
$$\mathrm{log}_{\sqrt{\mathrm{8}}} \left(\mathrm{6}×\mathrm{log}_{\mathrm{24}} \mathrm{6}−\mathrm{log}_{\mathrm{24}} \mathrm{3}\right) \\ $$$$\mathrm{FAILED}\:\mathrm{TO}\:\mathrm{CALCULATE} \\ $$$$\mathrm{why}\:\mathrm{do}\:\mathrm{tinku}\:\mathrm{tara}\:\mathrm{can}'\mathrm{t}\:\mathrm{do}\:\mathrm{log}_{{b}} {x} \\ $$