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Question Number 199389 by depressiveshrek last updated on 02/Nov/23

log_(12) 60=? log_6 30=a log_(15) 24=b

$$\mathrm{log}_{\mathrm{12}} \mathrm{60}=? \\ $$$$\mathrm{log}_{\mathrm{6}} \mathrm{30}={a} \\ $$$$\mathrm{log}_{\mathrm{15}} \mathrm{24}={b} \\ $$

Answered by cortano12 last updated on 03/Nov/23

log _(12) (60)=((log _6 (6.5.2))/(log _6 (6.2))) = ((2ab+2a−1)/(ab+b+1))

$$\:\mathrm{log}\:_{\mathrm{12}} \left(\mathrm{60}\right)=\frac{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{6}.\mathrm{5}.\mathrm{2}\right)}{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{6}.\mathrm{2}\right)} \\ $$$$\:\:=\:\frac{\mathrm{2ab}+\mathrm{2a}−\mathrm{1}}{\mathrm{ab}+\mathrm{b}+\mathrm{1}}\: \\ $$

Commented by depressiveshrek last updated on 03/Nov/23

I don′t understand how you got there...

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{how}\:\mathrm{you}\:\mathrm{got}\:\mathrm{there}... \\ $$

Answered by mr W last updated on 04/Nov/23

log_(15) 24=((log_6 24)/(log_6 15))=((log_6 6+2log_6 2)/(log_6 30−log_6 2)) =((1+2log_6 2)/(log_6 30−log_6 2)) =((1+2log_6 2)/(a−log_6 2))=b ⇒log_6 2=((ab−1)/(2+b)) log_(12) 60 =((log_6 30+log_6 2)/(log_6 6+log_6 2)) =((log_6 30+log_6 2)/(1+log_6 2)) =((a+((ab−1)/(2+b)))/(1+((ab−1)/(2+b)))) =((2a(b+1)−1)/((a+1)b+1)) ✓

$$\mathrm{log}_{\mathrm{15}} \:\mathrm{24}=\frac{\mathrm{log}_{\mathrm{6}} \:\mathrm{24}}{\mathrm{log}_{\mathrm{6}} \:\mathrm{15}}=\frac{\mathrm{log}_{\mathrm{6}} \:\mathrm{6}+\mathrm{2log}_{\mathrm{6}} \:\mathrm{2}}{\mathrm{log}_{\mathrm{6}} \:\mathrm{30}−\mathrm{log}_{\mathrm{6}} \:\mathrm{2}} \\ $$$$=\frac{\mathrm{1}+\mathrm{2log}_{\mathrm{6}} \:\mathrm{2}}{\mathrm{log}_{\mathrm{6}} \:\mathrm{30}−\mathrm{log}_{\mathrm{6}} \:\mathrm{2}} \\ $$$$=\frac{\mathrm{1}+\mathrm{2log}_{\mathrm{6}} \:\mathrm{2}}{{a}−\mathrm{log}_{\mathrm{6}} \:\mathrm{2}}={b} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{6}} \:\mathrm{2}=\frac{{ab}−\mathrm{1}}{\mathrm{2}+{b}} \\ $$$$ \\ $$$$\mathrm{log}_{\mathrm{12}} \:\mathrm{60} \\ $$$$=\frac{\mathrm{log}_{\mathrm{6}} \:\mathrm{30}+\mathrm{log}_{\mathrm{6}} \:\mathrm{2}}{\mathrm{log}_{\mathrm{6}} \:\mathrm{6}+\mathrm{log}_{\mathrm{6}} \:\mathrm{2}} \\ $$$$=\frac{\mathrm{log}_{\mathrm{6}} \:\mathrm{30}+\mathrm{log}_{\mathrm{6}} \:\mathrm{2}}{\mathrm{1}+\mathrm{log}_{\mathrm{6}} \:\mathrm{2}} \\ $$$$=\frac{{a}+\frac{{ab}−\mathrm{1}}{\mathrm{2}+{b}}}{\mathrm{1}+\frac{{ab}−\mathrm{1}}{\mathrm{2}+{b}}} \\ $$$$=\frac{\mathrm{2}{a}\left({b}+\mathrm{1}\right)−\mathrm{1}}{\left({a}+\mathrm{1}\right){b}+\mathrm{1}}\:\checkmark \\ $$

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