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Question Number 108240 by mathdave last updated on 15/Aug/20

Answered by john santu last updated on 15/Aug/20

((□JS△)/★) 2^(a+4) = 6 ⇒2^(a+4) = 2.3 ⇒2^(a+3) = 3 ...(i) 3^(2b+1) = 3.2^2 ⇒3^(2b) = 2^2 ; 3^b =2 then (2^(a+3) )^(2b) = 2^2 2^(2ab+6b) = 2^2 ⇒ab+3b = 1 ⇒ab−1 = −3b ; ab−2=−3b−1 we obtain 3^(ab−2) = 3^(−3b−1) =(1/((3^b )^3 ×3)) 3^(ab−2) = (1/(24)) .

$$\:\:\:\:\frac{\Box{JS}\bigtriangleup}{\bigstar} \\ $$$$\mathrm{2}^{{a}+\mathrm{4}} \:=\:\mathrm{6}\:\Rightarrow\mathrm{2}^{{a}+\mathrm{4}} \:=\:\mathrm{2}.\mathrm{3} \\ $$$$\Rightarrow\mathrm{2}^{{a}+\mathrm{3}} \:=\:\mathrm{3}\:...\left({i}\right) \\ $$$$\mathrm{3}^{\mathrm{2}{b}+\mathrm{1}} \:=\:\mathrm{3}.\mathrm{2}^{\mathrm{2}} \Rightarrow\mathrm{3}^{\mathrm{2}{b}} \:=\:\mathrm{2}^{\mathrm{2}} ;\:\mathrm{3}^{{b}} =\mathrm{2} \\ $$$${then}\:\left(\mathrm{2}^{{a}+\mathrm{3}} \right)^{\mathrm{2}{b}} \:=\:\mathrm{2}^{\mathrm{2}} \\ $$$$\mathrm{2}^{\mathrm{2}{ab}+\mathrm{6}{b}} \:=\:\mathrm{2}^{\mathrm{2}} \Rightarrow{ab}+\mathrm{3}{b}\:=\:\mathrm{1}\: \\ $$$$\Rightarrow{ab}−\mathrm{1}\:=\:−\mathrm{3}{b}\:;\:{ab}−\mathrm{2}=−\mathrm{3}{b}−\mathrm{1} \\ $$$${we}\:{obtain}\:\mathrm{3}^{{ab}−\mathrm{2}} \:=\:\mathrm{3}^{−\mathrm{3}{b}−\mathrm{1}} =\frac{\mathrm{1}}{\left(\mathrm{3}^{{b}} \right)^{\mathrm{3}} ×\mathrm{3}} \\ $$$$\mathrm{3}^{{ab}−\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{24}}\:. \\ $$

Answered by 1549442205PVT last updated on 16/Aug/20

2^(a+4) =6⇒(a+4)ln2=ln(2.3)=ln2+ln3 ⇒aln2=ln3−3ln2=ln(3/8)⇒a=((ln(3/8))/(ln2))(1) 3^(2b+1) =12⇒(2b+1)ln3=ln(3.4)⇔ 2bln3+ln3=ln3+ln4⇒2bln3=ln4 ⇒b=((ln4)/(2ln3))=((2ln2)/(2ln3))=((ln2)/(ln3))(2) From(1)(2)we get ab=((ln(3/8))/(ln2))×((ln2)/(ln3)) =((ln(3/8))/(ln3))=log_3 (3/8) ⇒ab−2=log_3 (3/8)−2=log_3 (3/8)−log_3 3^2 =log_3 ((3/8)/9)=log_3 ((1/(24))).Therefore 3^(ab−2) =3^(log_3 ((1/(24)))) =(1/(24))(since a^(log_a N) =N) Consequently we choose C

$$\mathrm{2}^{\mathrm{a}+\mathrm{4}} =\mathrm{6}\Rightarrow\left(\mathrm{a}+\mathrm{4}\right)\mathrm{ln2}=\mathrm{ln}\left(\mathrm{2}.\mathrm{3}\right)=\mathrm{ln2}+\mathrm{ln3} \\ $$$$\Rightarrow\mathrm{aln2}=\mathrm{ln3}−\mathrm{3ln2}=\mathrm{ln}\frac{\mathrm{3}}{\mathrm{8}}\Rightarrow\mathrm{a}=\frac{\mathrm{ln}\frac{\mathrm{3}}{\mathrm{8}}}{\mathrm{ln2}}\left(\mathrm{1}\right) \\ $$$$\mathrm{3}^{\mathrm{2b}+\mathrm{1}} =\mathrm{12}\Rightarrow\left(\mathrm{2b}+\mathrm{1}\right)\mathrm{ln3}=\mathrm{ln}\left(\mathrm{3}.\mathrm{4}\right)\Leftrightarrow \\ $$$$\mathrm{2bln3}+\mathrm{ln3}=\mathrm{ln3}+\mathrm{ln4}\Rightarrow\mathrm{2bln3}=\mathrm{ln4} \\ $$$$\Rightarrow\mathrm{b}=\frac{\mathrm{ln4}}{\mathrm{2ln3}}=\frac{\mathrm{2ln2}}{\mathrm{2ln3}}=\frac{\mathrm{ln2}}{\mathrm{ln3}}\left(\mathrm{2}\right) \\ $$$$\mathrm{From}\left(\mathrm{1}\right)\left(\mathrm{2}\right)\mathrm{we}\:\mathrm{get}\:\mathrm{ab}=\frac{\mathrm{ln}\frac{\mathrm{3}}{\mathrm{8}}}{\mathrm{ln2}}×\frac{\mathrm{ln2}}{\mathrm{ln3}} \\ $$$$=\frac{\mathrm{ln}\frac{\mathrm{3}}{\mathrm{8}}}{\mathrm{ln3}}=\mathrm{log}_{\mathrm{3}} \frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{ab}−\mathrm{2}=\mathrm{log}_{\mathrm{3}} \frac{\mathrm{3}}{\mathrm{8}}−\mathrm{2}=\mathrm{log}_{\mathrm{3}} \frac{\mathrm{3}}{\mathrm{8}}−\mathrm{log}_{\mathrm{3}} \mathrm{3}^{\mathrm{2}} \\ $$$$=\mathrm{log}_{\mathrm{3}} \left(\frac{\mathrm{3}}{\mathrm{8}}/\mathrm{9}\right)=\mathrm{log}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{24}}\right).\mathrm{Therefore} \\ $$$$\mathrm{3}^{\mathrm{ab}−\mathrm{2}} =\mathrm{3}^{\mathrm{log}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{24}}\right)} =\frac{\mathrm{1}}{\mathrm{24}}\left(\mathrm{since}\:\mathrm{a}^{\mathrm{log}_{\mathrm{a}} \mathrm{N}} =\mathrm{N}\right) \\ $$$$\mathrm{Consequently}\:\mathrm{we}\:\mathrm{choose}\:\mathrm{C} \\ $$

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