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Question Number 215840 by golsendro last updated on 19/Jan/25

log _(24) 3= a and log _(24) 6 = (b/6) log _(√8) (b−4a)= ?

$$\:\:\:\mathrm{log}\:_{\mathrm{24}} \:\mathrm{3}=\:{a}\:\mathrm{and}\:\mathrm{log}\:_{\mathrm{24}} \:\mathrm{6}\:=\:\frac{{b}}{\mathrm{6}} \\ $$$$\:\:\:\mathrm{log}\:_{\sqrt{\mathrm{8}}} \:\left({b}−\mathrm{4}{a}\right)=\:? \\ $$

Commented by oubiji last updated on 19/Jan/25

b−4a=log_(24) 6^6 −log_(24) 3^4 =log_(24) (6^6 /3^4 ) =log_(24) 24^2 = 2 then : log_(√8) (b−4a)=log_(√8) 2 =((ln2)/(ln(√8))) =((ln2)/((3/2)ln2)) =(2/3)

$$\:\:\:\:{b}−\mathrm{4}{a}=\mathrm{log}_{\mathrm{24}} \mathrm{6}^{\mathrm{6}} −{log}_{\mathrm{24}} \mathrm{3}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={log}_{\mathrm{24}} \frac{\mathrm{6}^{\mathrm{6}} }{\mathrm{3}^{\mathrm{4}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={log}_{\mathrm{24}} \mathrm{24}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2} \\ $$$$\:\:\:{then}\:: \\ $$$$\:\:\:\:\:{log}_{\sqrt{\mathrm{8}}} \left({b}−\mathrm{4}{a}\right)={log}_{\sqrt{\mathrm{8}}} \mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{ln}\mathrm{2}}{{ln}\sqrt{\mathrm{8}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{ln}\mathrm{2}}{\frac{\mathrm{3}}{\mathrm{2}}{ln}\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 20/Jan/25

please next time post your solution as “answer” not as a “comment”

$${please}\:{next}\:{time}\:\:{post}\:{your}\:{solution}\:{as}\: \\ $$$$``{answer}''\:{not}\:{as}\:{a}\:``{comment}'' \\ $$

Answered by A5T last updated on 19/Jan/25

b=log_(24) 6^6 ; 4a=log_(24) 3^4 b−4a=log_(24) ((6^6 /3^4 ))=log_(24) (24^2 )=2 ⇒log_(√8) (b−4a)⇒log_(√8) (2)=log_2^(3/2) (2)=(2/3)

$$\mathrm{b}=\mathrm{log}_{\mathrm{24}} \mathrm{6}^{\mathrm{6}} \:\:;\:\:\:\:\mathrm{4a}=\mathrm{log}_{\mathrm{24}} \mathrm{3}^{\mathrm{4}} \\ $$$$\mathrm{b}−\mathrm{4a}=\mathrm{log}_{\mathrm{24}} \left(\frac{\mathrm{6}^{\mathrm{6}} }{\mathrm{3}^{\mathrm{4}} }\right)=\mathrm{log}_{\mathrm{24}} \left(\mathrm{24}^{\mathrm{2}} \right)=\mathrm{2} \\ $$$$\Rightarrow\mathrm{log}_{\sqrt{\mathrm{8}}} \left(\mathrm{b}−\mathrm{4a}\right)\Rightarrow\mathrm{log}_{\sqrt{\mathrm{8}}} \left(\mathrm{2}\right)=\mathrm{log}_{\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} } \left(\mathrm{2}\right)=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Answered by MathematicalUser2357 last updated on 26/Jan/25

log_(√8) (6×log_(24) 6−log_(24) 3) FAILED TO CALCULATE why do tinku tara can′t do log_b x

$$\mathrm{log}_{\sqrt{\mathrm{8}}} \left(\mathrm{6}×\mathrm{log}_{\mathrm{24}} \mathrm{6}−\mathrm{log}_{\mathrm{24}} \mathrm{3}\right) \\ $$$$\mathrm{FAILED}\:\mathrm{TO}\:\mathrm{CALCULATE} \\ $$$$\mathrm{why}\:\mathrm{do}\:\mathrm{tinku}\:\mathrm{tara}\:\mathrm{can}'\mathrm{t}\:\mathrm{do}\:\mathrm{log}_{{b}} {x} \\ $$

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