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Question Number 193949 by sonukgindia last updated on 23/Jun/23

Answered by aleks041103 last updated on 23/Jun/23

$$8^x-8x=f\left(x\right)$$$$f{}^{\prime }\left(x\right)=ln\left(8\right)8^x-8;f{}^{\prime }\left(x_0\right)=0$$$$f{}^{″}\left(x\right)=ln{\left(8\right)}^2{8}^x>0$$$$\Rightarrow atx=x_{0}wehaveaminimum.$$$$ln\left(8\right)8^{x_0}=8\Rightarrow 8^{x_0}=\frac{8}{ln\left(8\right)},x_0=\frac{ln\left(8\right)-ln\left(ln\left(8\right)\right)}{ln\left(8\right)}$$$$\Rightarrow f\left(x_0\right)=\frac{8}{ln\left(8\right)}-8+\frac{8ln\left(ln\left(8\right)\right)}{ln\left(8\right)}=$$$$=\frac{8}{ln\left(8\right)}(1+ln\left(ln\left(8\right)\right)-ln\left(8\right))$$$$butln\left(ln\left(8\right)\right)<ln\left(8\right)-1\Rightarrow f\left(x_0\right)<0$$$$\Rightarrow \mathrm{\exists }{x}_1\ne x_2:f\left(x_1\right)=f\left(x_2\right)=0$$$$obv.oneis{x}_1=1,butthereisonemore$$$$whichis{x}_2\approx 0.183$$$$noexactsolutionunlessyouuselambertW$$

Answered by Subhi last updated on 24/Jun/23

$$1=8x.2^{-3x}$$$$1=8x.e^{-3xln\left(2\right)}$$$$-\frac{3}{8}ln\left(2\right)=-3xln\left(2\right).e^{-3xln\left(2\right)}$$$$fromW\left({xe}^x\right)=x$$$$W\left(\frac{-3}{8}ln\left(2\right)\right)=-3ln\left(2\right)x$$$$x=\frac{W\left(\frac{-3}{8}ln\left(2\right)\right)}{-3ln\left(2\right)}$$

Answered by Spillover last updated on 24/Jun/23

$$lety=2^{3x}=8xy=2^{3x}y=8x$$$$drawthegraphthenfindpointof$$$$intersection$$$$$$

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