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pre-calculus-if-log-48-72-log-54-12-k-then-log-12-27-

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Question Number 135285 by mnjuly1970 last updated on 11/Mar/21
pre−calculus if log_(48) ^(72) +log_(54) ^(12) =k then log_( 12) ^( 27) =???
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{pre}−{calculus} \\ $$$$\:\:{if}\:\:\:\:\:\:\:{log}_{\mathrm{48}} ^{\mathrm{72}} +{log}_{\mathrm{54}} ^{\mathrm{12}} \:={k}\: \\ $$$$\:\:\:\:\:\:{then}\:\:{log}_{\:\mathrm{12}} ^{\:\mathrm{27}} \:\:=??? \\ $$$$\:\:\:\:\:\:\:\: \\ $$
Answered by bobhans last updated on 12/Mar/21
((ln 72)/(ln 48)) + ((ln 12)/(ln 54)) = k ((ln 6+ln 12)/(ln 6+ln 8)) + ((ln 12)/(ln 9+ln 6)) = k (1)ln 6 = ln 2+ln 3=a+b (2)((3ln 2+2ln 3)/(4ln 2+ln 3)) + ((2ln 2+ln 3 )/(3ln 3+ln 2)) = k ⇒ ((3a+2b)/(4a+b)) + ((2a+b)/(a+3b)) = k ⇒(3a+2b)(a+3b)+(4a+b)(2a+b)=k(4a+b)(a+3b) ⇒3a^2 +11ab+6b^2 +8a^2 +6ab+b^2 =4ka^2 +13kab+3kb^2 ⇒(11−4k)a^2 +(17b−13kb)a+7b^2 −3kb^2 =0 ⇒a = ((13kb−17b + (√((17b−13kb)^2 −4(11−4k)((7b^2 −3kb^2 ))))/(22−8k)) log_(12) (27) = ((ln 27)/(ln 12)) = ((3ln 3)/(2ln 2+ln 3))=((3b)/(2a+b))
$$\:\frac{\mathrm{ln}\:\mathrm{72}}{\mathrm{ln}\:\mathrm{48}}\:+\:\frac{\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{54}}\:=\:{k} \\ $$$$\frac{\mathrm{ln}\:\mathrm{6}+\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{6}+\mathrm{ln}\:\mathrm{8}}\:+\:\frac{\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{9}+\mathrm{ln}\:\mathrm{6}}\:=\:{k} \\ $$$$\left(\mathrm{1}\right)\mathrm{ln}\:\mathrm{6}\:=\:\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3}={a}+{b} \\ $$$$\left(\mathrm{2}\right)\frac{\mathrm{3ln}\:\mathrm{2}+\mathrm{2ln}\:\mathrm{3}}{\mathrm{4ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3}}\:+\:\frac{\mathrm{2ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3}\:}{\mathrm{3ln}\:\mathrm{3}+\mathrm{ln}\:\mathrm{2}}\:=\:{k} \\ $$$$\Rightarrow\:\frac{\mathrm{3}{a}+\mathrm{2}{b}}{\mathrm{4}{a}+{b}}\:+\:\frac{\mathrm{2}{a}+{b}}{{a}+\mathrm{3}{b}}\:=\:{k}\: \\ $$$$\Rightarrow\left(\mathrm{3}{a}+\mathrm{2}{b}\right)\left({a}+\mathrm{3}{b}\right)+\left(\mathrm{4}{a}+{b}\right)\left(\mathrm{2}{a}+{b}\right)={k}\left(\mathrm{4}{a}+{b}\right)\left({a}+\mathrm{3}{b}\right) \\ $$$$\Rightarrow\mathrm{3}{a}^{\mathrm{2}} +\mathrm{11}{ab}+\mathrm{6}{b}^{\mathrm{2}} +\mathrm{8}{a}^{\mathrm{2}} +\mathrm{6}{ab}+{b}^{\mathrm{2}} =\mathrm{4}{ka}^{\mathrm{2}} +\mathrm{13}{kab}+\mathrm{3}{kb}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{11}−\mathrm{4}{k}\right){a}^{\mathrm{2}} +\left(\mathrm{17}{b}−\mathrm{13}{kb}\right){a}+\mathrm{7}{b}^{\mathrm{2}} −\mathrm{3}{kb}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}\:=\:\frac{\mathrm{13}{kb}−\mathrm{17}{b}\:+\:\sqrt{\left(\mathrm{17}{b}−\mathrm{13}{kb}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{11}−\mathrm{4}{k}\right)\left(\left(\mathrm{7}{b}^{\mathrm{2}} −\mathrm{3}{kb}^{\mathrm{2}} \right)\right.}}{\mathrm{22}−\mathrm{8}{k}} \\ $$$$ \\ $$$$\mathrm{log}_{\mathrm{12}} \left(\mathrm{27}\right)\:=\:\frac{\mathrm{ln}\:\mathrm{27}}{\mathrm{ln}\:\mathrm{12}}\:=\:\frac{\mathrm{3ln}\:\mathrm{3}}{\mathrm{2ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3}}=\frac{\mathrm{3}{b}}{\mathrm{2}{a}+{b}} \\ $$$$ \\ $$

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