Question-110050

Question Number 110050 by mathdave last updated on 26/Aug/20

Answered by Aziztisffola last updated on 27/Aug/20

9^x +6^x =4^x ⇒((9/4))^x +((6/4))^x =1 (((3/2))^2 )^x +((3/2))^x =1⇔((3/2))^(2x) +((3/2))^x =1 let u=((3/2))^x ⇒u^2 +u=1 u^2 +u−1=0 ⇒ △=1+4=5 u_1 =((−1+(√5))/2) >0 and u_2 =((−1−(√5))/2)<0 ⇒((3/2))^x =((−1+(√5))/2) ⇒ln(((3/2))^x )=ln(((−1+(√5))/2) ) xln((3/2))=ln(((−1+(√5))/2) ) x=((ln(((−1+(√5))/2) ))/(ln((3/2))))= log_(3/2) (((−1+(√5))/2)) something went wrong in the question the equation is 4^x +6^x =9^x this one gives us x=log_(3/2) (((1+(√5))/2))

$$\mathrm{9}^{\mathrm{x}} +\mathrm{6}^{\mathrm{x}} =\mathrm{4}^{\mathrm{x}} \Rightarrow\left(\frac{\mathrm{9}}{\mathrm{4}}\right)^{\mathrm{x}} +\left(\frac{\mathrm{6}}{\mathrm{4}}\right)^{\mathrm{x}} =\mathrm{1} \\ $$$$\left(\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \right)^{\mathrm{x}} +\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} =\mathrm{1}\Leftrightarrow\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2x}} +\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} =\mathrm{1} \\ $$$$\mathrm{let}\:\mathrm{u}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} \Rightarrow\mathrm{u}^{\mathrm{2}} +\mathrm{u}=\mathrm{1} \\ $$$$\:\mathrm{u}^{\mathrm{2}} +\mathrm{u}−\mathrm{1}=\mathrm{0}\:\Rightarrow\:\bigtriangleup=\mathrm{1}+\mathrm{4}=\mathrm{5} \\ $$$$\:\mathrm{u}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:>\mathrm{0}\:\:\mathrm{and}\:\mathrm{u}_{\mathrm{2}} =\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}<\mathrm{0}\: \\ $$$$\Rightarrow\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\mathrm{ln}\left(\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} \right)=\mathrm{ln}\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\right) \\ $$$$\:\mathrm{xln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{ln}\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\right) \\ $$$$\:\mathrm{x}=\frac{\mathrm{ln}\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\right)}{\mathrm{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}=\:\mathrm{log}_{\frac{\mathrm{3}}{\mathrm{2}}} \left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$\:\mathrm{something}\:\mathrm{went}\:\mathrm{wrong}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question} \\ $$$$\:\mathrm{the}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{4}^{\mathrm{x}} +\mathrm{6}^{\mathrm{x}} =\mathrm{9}^{\mathrm{x}} \\ $$$$\:\mathrm{this}\:\mathrm{one}\:\mathrm{gives}\:\mathrm{us}\:\mathrm{x}=\mathrm{log}_{\frac{\mathrm{3}}{\mathrm{2}}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$

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