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Question Number 165433 by mathlove last updated on 01/Feb/22

Answered by aleks041103 last updated on 01/Feb/22

(3/2^x )+(3/3^x )=2(((3/2)^x )/2^x )+2(((3/2)^(−x) )/3^x ) ⇒((3−2(3/2)^x )/2^x )=((2(3/2)^(−x) −3)/3^x ) ⇒((3/2))^x (3−2((3/2))^x )=2((3/2))^(−x) −3 t=((3/2))^x ⇒t(3−2t)=(2/t)−3 3t^2 −2t^3 +3t−2=0 2t^3 −3t^2 −3t+2=0 2(t^3 +1)−3t(t+1)=0 2(t+1)(t^2 −t+1)−3t(t+1)=0 (t+1)(2t^2 −5t+2)=0 t_1 =−1 t_(2,3) =((5±(√(25−4.2.2)))/4)=((5±3)/4)=(1/2);2 t=((3/2))^x ⇒x=log_(3/2) t=((log_2 t)/(log_2 3−1)) t≥0 x_1 =((log_2 (1/2))/(log_2 3−1))=(1/(1−log_2 3)) x_2 =((log_2 2)/(log_2 3−1))=(1/(log_2 3−1)) x=(1/(log_2 3−1));(1/(1−log_2 3))

$$ \\ $$$$\frac{\mathrm{3}}{\mathrm{2}^{{x}} }+\frac{\mathrm{3}}{\mathrm{3}^{{x}} }=\mathrm{2}\frac{\left(\mathrm{3}/\mathrm{2}\right)^{{x}} }{\mathrm{2}^{{x}} }+\mathrm{2}\frac{\left(\mathrm{3}/\mathrm{2}\right)^{−{x}} }{\mathrm{3}^{{x}} } \\ $$$$\Rightarrow\frac{\mathrm{3}−\mathrm{2}\left(\mathrm{3}/\mathrm{2}\right)^{{x}} }{\mathrm{2}^{{x}} }=\frac{\mathrm{2}\left(\mathrm{3}/\mathrm{2}\right)^{−{x}} −\mathrm{3}}{\mathrm{3}^{{x}} } \\ $$$$\Rightarrow\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} \left(\mathrm{3}−\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} \right)=\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{−{x}} −\mathrm{3} \\ $$$${t}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} \\ $$$$\Rightarrow{t}\left(\mathrm{3}−\mathrm{2}{t}\right)=\frac{\mathrm{2}}{{t}}−\mathrm{3} \\ $$$$\mathrm{3}{t}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{3}{t}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{2}{t}^{\mathrm{3}} −\mathrm{3}{t}^{\mathrm{2}} −\mathrm{3}{t}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{2}\left({t}^{\mathrm{3}} +\mathrm{1}\right)−\mathrm{3}{t}\left({t}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{2}\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)−\mathrm{3}{t}\left({t}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({t}+\mathrm{1}\right)\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{2}\right)=\mathrm{0} \\ $$$${t}_{\mathrm{1}} =−\mathrm{1} \\ $$$${t}_{\mathrm{2},\mathrm{3}} =\frac{\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{4}.\mathrm{2}.\mathrm{2}}}{\mathrm{4}}=\frac{\mathrm{5}\pm\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}};\mathrm{2} \\ $$$${t}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} \Rightarrow{x}={log}_{\mathrm{3}/\mathrm{2}} {t}=\frac{{log}_{\mathrm{2}} {t}}{{log}_{\mathrm{2}} \mathrm{3}−\mathrm{1}} \\ $$$${t}\geqslant\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\frac{{log}_{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}}{{log}_{\mathrm{2}} \mathrm{3}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{1}−{log}_{\mathrm{2}} \mathrm{3}} \\ $$$${x}_{\mathrm{2}} =\frac{{log}_{\mathrm{2}} \mathrm{2}}{{log}_{\mathrm{2}} \mathrm{3}−\mathrm{1}}=\frac{\mathrm{1}}{{log}_{\mathrm{2}} \mathrm{3}−\mathrm{1}} \\ $$$$ \\ $$$${x}=\frac{\mathrm{1}}{{log}_{\mathrm{2}} \mathrm{3}−\mathrm{1}};\frac{\mathrm{1}}{\mathrm{1}−{log}_{\mathrm{2}} \mathrm{3}} \\ $$

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