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Question Number 65934 by gunawan last updated on 06/Aug/19

log_7 2=a log_2 3=b log_6 98=...

$$\:\mathrm{log}_{\mathrm{7}} \:\mathrm{2}={a} \\ $$$$\mathrm{log}_{\mathrm{2}} \:\mathrm{3}={b} \\ $$$$\:\mathrm{log}_{\mathrm{6}} \:\mathrm{98}=... \\ $$

Answered by meme last updated on 06/Aug/19

answers log_7 2 = ((ln2)/(ln7)) = 0.35 = a log_2 3 = ((ln3)/(ln2)) = 1.58=b log_6 98=((ln98)/(ln6))= 2.55

$$\:\:\:\:\:\:\:\:\:\:\:\:{answers} \\ $$$$\:\:\:\:\:\:\:\:{log}_{\mathrm{7}} \mathrm{2}\:=\:\frac{{ln}\mathrm{2}}{{ln}\mathrm{7}}\:=\:\mathrm{0}.\mathrm{35}\:=\:{a} \\ $$$$\:\:\:\:\:\:\:\:{log}_{\mathrm{2}} \mathrm{3}\:=\:\frac{{ln}\mathrm{3}}{{ln}\mathrm{2}}\:=\:\mathrm{1}.\mathrm{58}={b} \\ $$$$\:\:\:\:\:\:\:\:{log}_{\mathrm{6}} \mathrm{98}=\frac{{ln}\mathrm{98}}{{ln}\mathrm{6}}=\:\mathrm{2}.\mathrm{55} \\ $$

Answered by Kunal12588 last updated on 06/Aug/19

log_6 98=log_6 (2×7^2 ) =log_6 2+2log_6 7 =((log_7 2)/(log_7 (2×3)))+((2log_2 7)/(log_2 2+log_2 3)) =(a/(a+log_7 3))+((2× ((log_7 7)/(log_7 2)))/(1+b)) =(a/(a+((log_2 3)/(log_2 7))))+(2/(a(1+b))) ⇒(a/(a+ab))+(2/(a+ab)) ⇒((a+2)/(a+ab))

$${log}_{\mathrm{6}} \mathrm{98}={log}_{\mathrm{6}} \left(\mathrm{2}×\mathrm{7}^{\mathrm{2}} \right) \\ $$$$={log}_{\mathrm{6}} \mathrm{2}+\mathrm{2}{log}_{\mathrm{6}} \mathrm{7} \\ $$$$=\frac{{log}_{\mathrm{7}} \mathrm{2}}{{log}_{\mathrm{7}} \left(\mathrm{2}×\mathrm{3}\right)}+\frac{\mathrm{2}{log}_{\mathrm{2}} \mathrm{7}}{{log}_{\mathrm{2}} \mathrm{2}+{log}_{\mathrm{2}} \mathrm{3}} \\ $$$$=\frac{{a}}{{a}+{log}_{\mathrm{7}} \mathrm{3}}+\frac{\mathrm{2}×\:\frac{{log}_{\mathrm{7}} \mathrm{7}}{{log}_{\mathrm{7}} \mathrm{2}}}{\mathrm{1}+{b}} \\ $$$$=\frac{{a}}{{a}+\frac{{log}_{\mathrm{2}} \mathrm{3}}{{log}_{\mathrm{2}} \mathrm{7}}}+\frac{\mathrm{2}}{{a}\left(\mathrm{1}+{b}\right)} \\ $$$$\Rightarrow\frac{{a}}{{a}+{ab}}+\frac{\mathrm{2}}{{a}+{ab}} \\ $$$$\Rightarrow\frac{{a}+\mathrm{2}}{{a}+{ab}} \\ $$

Answered by Kunal12588 last updated on 06/Aug/19

log_7 2=((ln 2)/(ln7))=a log_2 3=((ln 3)/(ln 2))=b log_6 98=((ln 2 +2 ln 7 )/(ln 2 + ln 3)) =((((ln 2)/(ln 2))+((2 ln 7)/(ln 2)))/(((ln 2)/(ln 2))+((ln 3)/(ln 2)))) =((1+(2/a))/(1+b)) =((a+2)/(a+ab))

$${log}_{\mathrm{7}} \:\mathrm{2}=\frac{{ln}\:\mathrm{2}}{{ln}\mathrm{7}}={a} \\ $$$${log}_{\mathrm{2}} \:\mathrm{3}=\frac{{ln}\:\mathrm{3}}{{ln}\:\mathrm{2}}={b} \\ $$$${log}_{\mathrm{6}} \mathrm{98}=\frac{{ln}\:\mathrm{2}\:+\mathrm{2}\:{ln}\:\mathrm{7}\:}{{ln}\:\mathrm{2}\:+\:{ln}\:\mathrm{3}} \\ $$$$=\frac{\frac{{ln}\:\mathrm{2}}{{ln}\:\mathrm{2}}+\frac{\mathrm{2}\:{ln}\:\mathrm{7}}{{ln}\:\mathrm{2}}}{\frac{{ln}\:\mathrm{2}}{{ln}\:\mathrm{2}}+\frac{{ln}\:\mathrm{3}}{{ln}\:\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}+\frac{\mathrm{2}}{{a}}}{\mathrm{1}+{b}} \\ $$$$=\frac{{a}+\mathrm{2}}{{a}+{ab}} \\ $$

Commented by gunawan last updated on 06/Aug/19

thank you Sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

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