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Question Number 123845 by Akeyz last updated on 28/Nov/20

Commented by TANMAY PANACEA last updated on 28/Nov/20

n=1 ((8+27)/(12+18))=((35)/(30))=(7/6)

$${n}=\mathrm{1} \\ $$$$\frac{\mathrm{8}+\mathrm{27}}{\mathrm{12}+\mathrm{18}}=\frac{\mathrm{35}}{\mathrm{30}}=\frac{\mathrm{7}}{\mathrm{6}} \\ $$

Answered by floor(10²Eta[1]) last updated on 28/Nov/20

6.(2^(3n) +3^(3n) )=7(2^(2n) .3^n +2^n .3^(2n) ) 2^n =a, 3^n =b 6(a^3 +b^3 )=7(a^2 b+ab^2 ) 6(a+b)(a^2 −ab+b^2 )=7ab(a+b) 6(a^2 +b^2 −ab)=7ab 6a^2 −13ab+6b^2 =0 a=((13b±5b)/(12))=((3b)/2), ((2b)/3) 2^n =((3.3^n )/2)⇒2^(n+1) =3^(n+1) ⇒n=−1 2^n =((2.3^n )/3)⇒2^(n−1) =3^(n−1) ⇒n=1

$$\mathrm{6}.\left(\mathrm{2}^{\mathrm{3n}} +\mathrm{3}^{\mathrm{3n}} \right)=\mathrm{7}\left(\mathrm{2}^{\mathrm{2n}} .\mathrm{3}^{\mathrm{n}} +\mathrm{2}^{\mathrm{n}} .\mathrm{3}^{\mathrm{2n}} \right) \\ $$$$\mathrm{2}^{\mathrm{n}} =\mathrm{a},\:\mathrm{3}^{\mathrm{n}} =\mathrm{b} \\ $$$$\mathrm{6}\left(\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} \right)=\mathrm{7}\left(\mathrm{a}^{\mathrm{2}} \mathrm{b}+\mathrm{ab}^{\mathrm{2}} \right) \\ $$$$\mathrm{6}\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}^{\mathrm{2}} −\mathrm{ab}+\mathrm{b}^{\mathrm{2}} \right)=\mathrm{7ab}\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\mathrm{6}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{ab}\right)=\mathrm{7ab} \\ $$$$\mathrm{6a}^{\mathrm{2}} −\mathrm{13ab}+\mathrm{6b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{a}=\frac{\mathrm{13b}\pm\mathrm{5b}}{\mathrm{12}}=\frac{\mathrm{3b}}{\mathrm{2}},\:\frac{\mathrm{2b}}{\mathrm{3}}\: \\ $$$$\mathrm{2}^{\mathrm{n}} =\frac{\mathrm{3}.\mathrm{3}^{\mathrm{n}} }{\mathrm{2}}\Rightarrow\mathrm{2}^{\mathrm{n}+\mathrm{1}} =\mathrm{3}^{\mathrm{n}+\mathrm{1}} \Rightarrow\mathrm{n}=−\mathrm{1} \\ $$$$\mathrm{2}^{\mathrm{n}} =\frac{\mathrm{2}.\mathrm{3}^{\mathrm{n}} }{\mathrm{3}}\Rightarrow\mathrm{2}^{\mathrm{n}−\mathrm{1}} =\mathrm{3}^{\mathrm{n}−\mathrm{1}} \Rightarrow\mathrm{n}=\mathrm{1} \\ $$

Commented by WilliamsErinfolami last updated on 28/Nov/20

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