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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(x) f ′(x)=ln(8)8^x −8; f ′(x_0 )=0 f ′′(x)=ln(8)^2 8^x >0 ⇒at x=x_0 we have a minimum. ln(8)8^x_0 =8⇒8^x_0 =(8/(ln(8))), x_0 =((ln(8)−ln(ln(8)))/(ln(8))) ⇒f(x_0 )=(8/(ln(8)))−8+((8ln(ln(8)))/(ln(8)))= =(8/(ln(8)))(1+ln(ln(8))−ln(8)) but ln(ln(8))<ln(8)−1⇒f(x_0 )<0 ⇒∃x_1 ≠x_2 : f(x_1 )=f(x_2 )=0 obv. one is x_1 =1, but there is one more which is x_2 ≈0.183 no exact solution unless you use lambert W

$$\mathrm{8}^{{x}} −\mathrm{8}{x}={f}\left({x}\right) \\ $$$${f}\:'\left({x}\right)={ln}\left(\mathrm{8}\right)\mathrm{8}^{{x}} −\mathrm{8};\:{f}\:'\left({x}_{\mathrm{0}} \right)=\mathrm{0} \\ $$$${f}\:''\left({x}\right)={ln}\left(\mathrm{8}\right)^{\mathrm{2}} \mathrm{8}^{{x}} >\mathrm{0} \\ $$$$\Rightarrow{at}\:{x}={x}_{\mathrm{0}} \:{we}\:{have}\:{a}\:{minimum}. \\ $$$${ln}\left(\mathrm{8}\right)\mathrm{8}^{{x}_{\mathrm{0}} } =\mathrm{8}\Rightarrow\mathrm{8}^{{x}_{\mathrm{0}} } =\frac{\mathrm{8}}{{ln}\left(\mathrm{8}\right)},\:{x}_{\mathrm{0}} =\frac{{ln}\left(\mathrm{8}\right)−{ln}\left({ln}\left(\mathrm{8}\right)\right)}{{ln}\left(\mathrm{8}\right)} \\ $$$$\Rightarrow{f}\left({x}_{\mathrm{0}} \right)=\frac{\mathrm{8}}{{ln}\left(\mathrm{8}\right)}−\mathrm{8}+\frac{\mathrm{8}{ln}\left({ln}\left(\mathrm{8}\right)\right)}{{ln}\left(\mathrm{8}\right)}= \\ $$$$=\frac{\mathrm{8}}{{ln}\left(\mathrm{8}\right)}\left(\mathrm{1}+{ln}\left({ln}\left(\mathrm{8}\right)\right)−{ln}\left(\mathrm{8}\right)\right) \\ $$$${but}\:{ln}\left({ln}\left(\mathrm{8}\right)\right)<{ln}\left(\mathrm{8}\right)−\mathrm{1}\Rightarrow{f}\left({x}_{\mathrm{0}} \right)<\mathrm{0} \\ $$$$\Rightarrow\exists{x}_{\mathrm{1}} \neq{x}_{\mathrm{2}} :\:{f}\left({x}_{\mathrm{1}} \right)={f}\left({x}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$${obv}.\:{one}\:{is}\:{x}_{\mathrm{1}} =\mathrm{1},\:{but}\:{there}\:{is}\:{one}\:{more} \\ $$$${which}\:{is}\:{x}_{\mathrm{2}} \approx\mathrm{0}.\mathrm{183} \\ $$$${no}\:{exact}\:{solution}\:{unless}\:{you}\:{use}\:{lambert}\:{W} \\ $$

Answered by Subhi last updated on 24/Jun/23

1 = 8x.2^(−3x) 1 = 8x.e^(−3xln(2)) −(3/8)ln(2) = −3xln(2).e^(−3xln(2)) from W(xe^x )=x W(((−3)/8)ln(2))=−3ln(2)x x = ((W(((−3)/8)ln(2)))/(−3ln(2)))

$$\mathrm{1}\:=\:\mathrm{8}{x}.\mathrm{2}^{−\mathrm{3}{x}} \:\: \\ $$$$\mathrm{1}\:=\:\mathrm{8}{x}.{e}^{−\mathrm{3}{xln}\left(\mathrm{2}\right)} \\ $$$$−\frac{\mathrm{3}}{\mathrm{8}}{ln}\left(\mathrm{2}\right)\:=\:−\mathrm{3}{xln}\left(\mathrm{2}\right).{e}^{−\mathrm{3}{xln}\left(\mathrm{2}\right)} \\ $$$${from}\:{W}\left({xe}^{{x}} \right)={x} \\ $$$${W}\left(\frac{−\mathrm{3}}{\mathrm{8}}{ln}\left(\mathrm{2}\right)\right)=−\mathrm{3}{ln}\left(\mathrm{2}\right){x} \\ $$$${x}\:=\:\frac{{W}\left(\frac{−\mathrm{3}}{\mathrm{8}}{ln}\left(\mathrm{2}\right)\right)}{−\mathrm{3}{ln}\left(\mathrm{2}\right)} \\ $$

Answered by Spillover last updated on 24/Jun/23

let y=2^(3x) =8x y=2^(3x ) y=8x draw the graph then find point of intersection

$${let}\:{y}=\mathrm{2}^{\mathrm{3}{x}} =\mathrm{8}{x}\:\:\:\:\:{y}=\mathrm{2}^{\mathrm{3}{x}\:\:} \:\:{y}=\mathrm{8}{x} \\ $$$${draw}\:{the}\:{graph}\:{then}\:{find}\:{point}\:{of}\: \\ $$$${intersection} \\ $$$$ \\ $$

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