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

Answered by mahdipoor last updated on 14/Aug/24

$$1)e^x=t$$$$t^3-3t-\frac{4}{t}=0\Rightarrow t^4-3t^2-4=0\Rightarrow$$$$t^2=\frac{3\pm \sqrt{9+16}}{2}=4,-1\Rightarrow e^x=t=\pm 2,\pm i$$$$x=ln2,\pm \frac{\pi }{2}i,ln2+\pi i$$$$2)2^{x-3}=t$$$$t^2\times 2^{+1}-3t+1=0\Rightarrow \frac{2^x}{8}=t=\frac{1}{4}(3\pm \sqrt{9-8)}=0.5,1$$$$x=3,2$$

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

$$e^x(e^{3x}-3e^x-4e^{-x}=0)$$$$e^{4x}-3e^{2x}-4=0$$$$e^{2x}=u$$$$u^2-3u-4=0$$$$(u-4)(u+1)=0$$$$u=4,-1$$$$e^{2x}=2^2,e^{2x}=-1$$$$e^x=\pm 2,e^x=\pm \mathit{i}$$$$x\ln e=\ln (\pm 2),x\ln e=\ln (\pm \mathit{i})$$$$x=\ln (\pm 2),x=\ln (\pm \mathit{i})$$

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

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

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