log-6-2-a-log-6-9-

Question Number 197249 by sciencestudentW last updated on 11/Sep/23

$$\mathrm{lo}\underset{\mathrm{6}} {\mathrm{g}2}={a} \\ $$$$\mathrm{lo}\underset{\mathrm{6}} {\mathrm{g}9}=?\:\: \\ $$

Answered by AST last updated on 11/Sep/23

((log2)/(log2+log3))=a⇒log2(1−a)=alog3 ((2log3)/(log2+log3))=(2/(((log2)/(log3))+1))=(2/((a/(1−a))+1))=((2/1)/(1/(1−a)))=2−2a

$$\frac{{log}\mathrm{2}}{{log}\mathrm{2}+{log}\mathrm{3}}={a}\Rightarrow{log}\mathrm{2}\left(\mathrm{1}−{a}\right)={alog}\mathrm{3} \\ $$$$\frac{\mathrm{2}{log}\mathrm{3}}{{log}\mathrm{2}+{log}\mathrm{3}}=\frac{\mathrm{2}}{\frac{{log}\mathrm{2}}{{log}\mathrm{3}}+\mathrm{1}}=\frac{\mathrm{2}}{\frac{{a}}{\mathrm{1}−{a}}+\mathrm{1}}=\frac{\frac{\mathrm{2}}{\mathrm{1}}}{\frac{\mathrm{1}}{\mathrm{1}−{a}}}=\mathrm{2}−\mathrm{2}{a} \\ $$

Answered by Rasheed.Sindhi last updated on 13/Sep/23

log_6 2=a; log_6 9=? log_6 2=a⇒log_6 ((6/3))=a log_6 6−log_6 3=a 1−log_6 3=a log_6 3=1−a ▶log_6 9=log_6 3^2 =2log_6 3=2(1−a)

$$\mathrm{log}_{\mathrm{6}} \mathrm{2}={a};\:\mathrm{log}_{\mathrm{6}} \mathrm{9}=? \\ $$$$\mathrm{log}_{\mathrm{6}} \mathrm{2}={a}\Rightarrow\mathrm{log}_{\mathrm{6}} \left(\frac{\mathrm{6}}{\mathrm{3}}\right)={a} \\ $$$$\mathrm{log}_{\mathrm{6}} \mathrm{6}−\mathrm{log}_{\mathrm{6}} \mathrm{3}={a}\:\:\:\:\: \\ $$$$\mathrm{1}−\mathrm{log}_{\mathrm{6}} \mathrm{3}={a}\:\:\:\:\: \\ $$$$\mathrm{log}_{\mathrm{6}} \mathrm{3}=\mathrm{1}−{a} \\ $$$$\blacktriangleright\mathrm{log}_{\mathrm{6}} \mathrm{9}=\mathrm{log}_{\mathrm{6}} \mathrm{3}^{\mathrm{2}} \:=\mathrm{2log}_{\mathrm{6}} \mathrm{3}=\mathrm{2}\left(\mathrm{1}−{a}\right)\:\: \\ $$

Answered by Rasheed.Sindhi last updated on 13/Sep/23

log_6 2=a;log_6 9=? a=log_6 2 =(1/2)log_6 4=(1/2)log_6 (((36)/9)) =(1/2){log_6 36−log_6 9} =(1/2){2−log_6 9}=a =1−((log_6 9)/2)=a log_6 9=2(1−a)

$$\mathrm{log}_{\mathrm{6}} \mathrm{2}={a};\mathrm{log}_{\mathrm{6}} \mathrm{9}=?\:\: \\ $$$${a}=\mathrm{log}_{\mathrm{6}} \mathrm{2} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{\mathrm{6}} \mathrm{4}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{\mathrm{6}} \left(\frac{\mathrm{36}}{\mathrm{9}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{log}_{\mathrm{6}} \mathrm{36}−\mathrm{log}_{\mathrm{6}} \mathrm{9}\right\}\:\:\:\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}−\mathrm{log}_{\mathrm{6}} \mathrm{9}\right\}={a} \\ $$$$=\mathrm{1}−\frac{\mathrm{log}_{\mathrm{6}} \mathrm{9}}{\mathrm{2}}={a} \\ $$$$\mathrm{log}_{\mathrm{6}} \mathrm{9}=\mathrm{2}\left(\mathrm{1}−{a}\right) \\ $$

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