Question-210643

Question Number 210643 by ChantalYah last updated on 14/Aug/24

Answered by mahdipoor last updated on 14/Aug/24

1)e^x =t t^3 −3t−(4/t)=0⇒t^4 −3t^2 −4=0⇒ t^2 =((3±(√(9+16)))/2)=4,−1⇒e^x =t=±2,±i x=ln2,±(π/2)i,ln2+πi 2)2^(x−3) =t t^2 ×2^(+1) −3t+1=0⇒(2^x /8)=t=(1/4)(3±(√(9−8)))=0.5,1 x=3,2

$$\left.\mathrm{1}\right){e}^{{x}} ={t} \\ $$$${t}^{\mathrm{3}} −\mathrm{3}{t}−\frac{\mathrm{4}}{{t}}=\mathrm{0}\Rightarrow{t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}=\mathrm{0}\Rightarrow \\ $$$${t}^{\mathrm{2}} =\frac{\mathrm{3}\pm\sqrt{\mathrm{9}+\mathrm{16}}}{\mathrm{2}}=\mathrm{4},−\mathrm{1}\Rightarrow{e}^{{x}} ={t}=\pm\mathrm{2},\pm{i} \\ $$$${x}={ln}\mathrm{2},\pm\frac{\pi}{\mathrm{2}}{i},{ln}\mathrm{2}+\pi{i} \\ $$$$\left.\mathrm{2}\right)\mathrm{2}^{{x}−\mathrm{3}} ={t} \\ $$$${t}^{\mathrm{2}} ×\mathrm{2}^{+\mathrm{1}} −\mathrm{3}{t}+\mathrm{1}=\mathrm{0}\Rightarrow\frac{\mathrm{2}^{{x}} }{\mathrm{8}}={t}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}\pm\sqrt{\left.\mathrm{9}−\mathrm{8}\right)}=\mathrm{0}.\mathrm{5},\mathrm{1}\right. \\ $$$${x}=\mathrm{3},\mathrm{2} \\ $$

Answered by Rasheed.Sindhi last updated on 14/Aug/24

e^x (e^(3x) −3e^x −4e^(−x) =0) e^(4x) −3e^(2x) −4=0 e^(2x) =u u^2 −3u−4=0 (u−4)(u+1)=0 u=4 , −1 e^(2x) =2^2 , e^(2x) =−1 e^x =±2 , e^x =±i xln e=ln(± 2) , xln e=ln(±i) x=ln(± 2) , x=ln(±i)

$${e}^{{x}} \left({e}^{\mathrm{3}{x}} −\mathrm{3}{e}^{{x}} −\mathrm{4}{e}^{−{x}} =\mathrm{0}\right) \\ $$$$\:\:\:\:{e}^{\mathrm{4}{x}} −\mathrm{3}{e}^{\mathrm{2}{x}} −\mathrm{4}=\mathrm{0} \\ $$$${e}^{\mathrm{2}{x}} ={u} \\ $$$$\:\:\:{u}^{\mathrm{2}} −\mathrm{3}{u}−\mathrm{4}=\mathrm{0} \\ $$$$\:\:\left({u}−\mathrm{4}\right)\left({u}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:{u}=\mathrm{4}\:,\:−\mathrm{1} \\ $$$$\:\:\:{e}^{\mathrm{2}{x}} =\mathrm{2}^{\mathrm{2}} \:\:,\:{e}^{\mathrm{2}{x}} =−\mathrm{1} \\ $$$$\:\:\:\:{e}^{{x}} =\pm\mathrm{2}\:,\:{e}^{{x}} =\pm\boldsymbol{{i}} \\ $$$$\:\:{x}\mathrm{ln}\:{e}=\mathrm{ln}\left(\pm\:\mathrm{2}\right)\:,\:{x}\mathrm{ln}\:{e}=\mathrm{ln}\left(\pm\boldsymbol{{i}}\right) \\ $$$$\:\:\:{x}=\mathrm{ln}\left(\pm\:\mathrm{2}\right)\:\:\:,\:\:{x}=\mathrm{ln}\left(\pm\boldsymbol{{i}}\right) \\ $$

Answered by Rasheed.Sindhi last updated on 14/Aug/24

(2^(2x) /2^5 )−3((2^x /2^3 ))+1=0 2^x =u (u^2 /4)−3u+8=0 u^2 −12u+32=0 (u−4)(u−8)=0 u=4,8 2^x =2^2 ,2^3 x=2,3

$$\frac{\mathrm{2}^{\mathrm{2}{x}} }{\mathrm{2}^{\mathrm{5}} }−\mathrm{3}\left(\frac{\mathrm{2}^{{x}} }{\mathrm{2}^{\mathrm{3}} }\right)+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2}^{{x}} ={u} \\ $$$$\frac{{u}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{3}{u}+\mathrm{8}=\mathrm{0} \\ $$$${u}^{\mathrm{2}} −\mathrm{12}{u}+\mathrm{32}=\mathrm{0} \\ $$$$\left({u}−\mathrm{4}\right)\left({u}−\mathrm{8}\right)=\mathrm{0} \\ $$$${u}=\mathrm{4},\mathrm{8} \\ $$$$\mathrm{2}^{{x}} =\mathrm{2}^{\mathrm{2}} ,\mathrm{2}^{\mathrm{3}} \\ $$$${x}=\mathrm{2},\mathrm{3} \\ $$

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