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Solve-for-x-8-x-27-x-12-x-18-x-7-6-




Question Number 77805 by Dah Solu Tion last updated on 10/Jan/20
Solve for x  ((8^x  + 27^x )/(12^x  + 18^x )) = (7/6)
$${Solve}\:{for}\:{x} \\ $$$$\frac{\mathrm{8}^{{x}} \:+\:\mathrm{27}^{{x}} }{\mathrm{12}^{{x}} \:+\:\mathrm{18}^{{x}} }\:=\:\frac{\mathrm{7}}{\mathrm{6}} \\ $$
Answered by john santu last updated on 10/Jan/20
12^x +18^x = 3^x ×(2^x )^2 +2^x ×(3^x )^2   let 2^x =t ,3^x =y  t^3 +y^3 =(7/6)(t^2 y+ty^2 )  6(t+y)(t^2 −ty+y^2 )=7ty(t+y)  6(t^2 −ty+y)^2 =7ty  6t^2 −13ty+6y^2 =0  (2t−3y)(3t−2y)=0  (i) 2t=3y ⇒2^(x+1) =3^(x+1) ⇒x=−1  (ii) 3×2^x =2×3^x  ⇒(3/2)=((3/2))^x ⇒x=1
$$\mathrm{12}^{{x}} +\mathrm{18}^{{x}} =\:\mathrm{3}^{{x}} ×\left(\mathrm{2}^{{x}} \right)^{\mathrm{2}} +\mathrm{2}^{{x}} ×\left(\mathrm{3}^{{x}} \right)^{\mathrm{2}} \\ $$$${let}\:\mathrm{2}^{{x}} ={t}\:,\mathrm{3}^{{x}} ={y} \\ $$$${t}^{\mathrm{3}} +{y}^{\mathrm{3}} =\frac{\mathrm{7}}{\mathrm{6}}\left({t}^{\mathrm{2}} {y}+{ty}^{\mathrm{2}} \right) \\ $$$$\mathrm{6}\left({t}+{y}\right)\left({t}^{\mathrm{2}} −{ty}+{y}^{\mathrm{2}} \right)=\mathrm{7}{ty}\left({t}+{y}\right) \\ $$$$\mathrm{6}\left({t}^{\mathrm{2}} −{ty}+{y}\right)^{\mathrm{2}} =\mathrm{7}{ty} \\ $$$$\mathrm{6}{t}^{\mathrm{2}} −\mathrm{13}{ty}+\mathrm{6}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{2}{t}−\mathrm{3}{y}\right)\left(\mathrm{3}{t}−\mathrm{2}{y}\right)=\mathrm{0} \\ $$$$\left({i}\right)\:\mathrm{2}{t}=\mathrm{3}{y}\:\Rightarrow\mathrm{2}^{{x}+\mathrm{1}} =\mathrm{3}^{{x}+\mathrm{1}} \Rightarrow{x}=−\mathrm{1} \\ $$$$\left({ii}\right)\:\mathrm{3}×\mathrm{2}^{{x}} =\mathrm{2}×\mathrm{3}^{{x}} \:\Rightarrow\frac{\mathrm{3}}{\mathrm{2}}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} \Rightarrow{x}=\mathrm{1} \\ $$
Commented by Dah Solu Tion last updated on 10/Jan/20
Thank boss
$${Thank}\:{boss} \\ $$

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