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Question-198413




Question Number 198413 by mokys last updated on 19/Oct/23
Commented by mokys last updated on 19/Oct/23
find log_(12) 60
$${find}\:{log}_{\mathrm{12}} \mathrm{60} \\ $$
Answered by cortano12 last updated on 19/Oct/23
   { ((log _(15) (24)=b⇒log _(24) (15)=(1/b))),((log _6 (30)=a⇒1+log _6 (5)=a)) :}    ((log _6 15)/(log _6 24)) =(1/b) ⇒((a−1+log _6 3)/(1+2log _6 2))=(1/b)    ((a−log _6 2)/(1+2log _6 2)) = (1/b) ⇒log _6 2 = ((ab−1)/(b+2))       log _(12) 60 = ((1+log _6 5+log _6 2)/(1+log _6 2))    = ((1+a−1+((ab−1)/(b+2)))/(1+((ab−1)/(b+2))))   = ((2ab+2a−1)/(ab+b+1))
$$\:\:\begin{cases}{\mathrm{log}\:_{\mathrm{15}} \left(\mathrm{24}\right)=\mathrm{b}\Rightarrow\mathrm{log}\:_{\mathrm{24}} \left(\mathrm{15}\right)=\frac{\mathrm{1}}{\mathrm{b}}}\\{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{30}\right)=\mathrm{a}\Rightarrow\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{5}\right)=\mathrm{a}}\end{cases} \\ $$$$\:\:\frac{\mathrm{log}\:_{\mathrm{6}} \mathrm{15}}{\mathrm{log}\:_{\mathrm{6}} \mathrm{24}}\:=\frac{\mathrm{1}}{\mathrm{b}}\:\Rightarrow\frac{\mathrm{a}−\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \mathrm{3}}{\mathrm{1}+\mathrm{2log}\:_{\mathrm{6}} \mathrm{2}}=\frac{\mathrm{1}}{\mathrm{b}} \\ $$$$\:\:\frac{\mathrm{a}−\mathrm{log}\:_{\mathrm{6}} \mathrm{2}}{\mathrm{1}+\mathrm{2log}\:_{\mathrm{6}} \mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{b}}\:\Rightarrow\mathrm{log}\:_{\mathrm{6}} \mathrm{2}\:=\:\frac{\mathrm{ab}−\mathrm{1}}{\mathrm{b}+\mathrm{2}} \\ $$$$\: \\ $$$$\:\:\mathrm{log}\:_{\mathrm{12}} \mathrm{60}\:=\:\frac{\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \mathrm{5}+\mathrm{log}\:_{\mathrm{6}} \mathrm{2}}{\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \mathrm{2}} \\ $$$$\:\:=\:\frac{\mathrm{1}+\mathrm{a}−\mathrm{1}+\frac{\mathrm{ab}−\mathrm{1}}{\mathrm{b}+\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{ab}−\mathrm{1}}{\mathrm{b}+\mathrm{2}}} \\ $$$$\:=\:\frac{\mathrm{2ab}+\mathrm{2a}−\mathrm{1}}{\mathrm{ab}+\mathrm{b}+\mathrm{1}}\: \\ $$$$ \\ $$$$\: \\ $$

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