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Question Number 12446 by tawa last updated on 22/Apr/17
find x (√x) = 8^x
$$\mathrm{find}\:\mathrm{x} \\ $$$$\sqrt{\mathrm{x}}\:\:=\:\:\mathrm{8}^{\mathrm{x}} \\ $$
Commented by mrW1 last updated on 23/Apr/17
For equation (√x)=a^x the solution is x=−((W(−2ln a))/(2ln a)) where 2ln a≤(1/e) or a≤e^(1/(2e)) ≈1.202 for a=8>1.202 ⇒ there is no solution.
$${For}\:{equation}\:\sqrt{{x}}={a}^{{x}} \:{the}\:{solution}\:{is} \\ $$$${x}=−\frac{{W}\left(−\mathrm{2ln}\:{a}\right)}{\mathrm{2ln}\:{a}} \\ $$$${where}\:\mathrm{2ln}\:{a}\leqslant\frac{\mathrm{1}}{{e}}\:{or} \\ $$$${a}\leqslant{e}^{\frac{\mathrm{1}}{\mathrm{2}{e}}} \approx\mathrm{1}.\mathrm{202} \\ $$$$ \\ $$$${for}\:{a}=\mathrm{8}>\mathrm{1}.\mathrm{202}\:\Rightarrow\:{there}\:{is}\:{no}\:{solution}. \\ $$
Answered by mrW1 last updated on 23/Apr/17
x≥0 x=^! 8^(2x) =64^x f(x)=x g(x)=64^x f(0)=0 g(0)=1>f(0) f′(x)=1 g′(x)=64^x ×ln 64>1=f′(x) ⇒for all x≥0 we have g(x)>f(x) i.e. there is no solution for f(x)=g(x)!
$${x}\geqslant\mathrm{0} \\ $$$${x}\overset{!} {=}\mathrm{8}^{\mathrm{2}{x}} =\mathrm{64}^{{x}} \\ $$$$ \\ $$$${f}\left({x}\right)={x} \\ $$$${g}\left({x}\right)=\mathrm{64}^{{x}} \\ $$$$ \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${g}\left(\mathrm{0}\right)=\mathrm{1}>{f}\left(\mathrm{0}\right) \\ $$$$ \\ $$$${f}'\left({x}\right)=\mathrm{1} \\ $$$${g}'\left({x}\right)=\mathrm{64}^{{x}} ×\mathrm{ln}\:\mathrm{64}>\mathrm{1}={f}'\left({x}\right) \\ $$$$ \\ $$$$\Rightarrow{for}\:{all}\:{x}\geqslant\mathrm{0}\:{we}\:{have}\:{g}\left({x}\right)>{f}\left({x}\right) \\ $$$${i}.{e}. \\ $$$${there}\:{is}\:{no}\:{solution}\:{for}\:{f}\left({x}\right)={g}\left({x}\right)! \\ $$
Commented by mrW1 last updated on 22/Apr/17
Commented by tawa last updated on 23/Apr/17
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$$$ \\ $$
Commented by geovane10math last updated on 23/Apr/17
The solution ∉ R ∈ C
$$\mathrm{The}\:\mathrm{solution}\:\notin\:\mathbb{R} \\ $$$$\in\:\mathbb{C} \\ $$
Answered by geovane10math last updated on 22/Apr/17
x = 8^(2x) x = 64^x x = e^(x∙ln 64) (x/e^(x∙ln 64) ) = 1 x∙e^(−x∙ln 64) = 1 (−ln 64)(x∙e^(−ln 64∙x) ) = −ln 64 −ln 64∙x∙e^(−ln 64∙x) = −ln 64 −ln 64∙x = y ye^y = −ln 64 y = W(−ln 64) W : Lambert function x = − ((W(−ln 64))/(ln 64))
$${x}\:=\:\mathrm{8}^{\mathrm{2}{x}} \\ $$$${x}\:=\:\mathrm{64}^{{x}} \\ $$$${x}\:=\:{e}^{{x}\centerdot\mathrm{ln}\:\mathrm{64}} \\ $$$$\frac{{x}}{{e}^{{x}\centerdot\mathrm{ln}\:\mathrm{64}} }\:=\:\mathrm{1} \\ $$$${x}\centerdot{e}^{−{x}\centerdot\mathrm{ln}\:\mathrm{64}} \:=\:\mathrm{1} \\ $$$$\left(−\mathrm{ln}\:\mathrm{64}\right)\left({x}\centerdot{e}^{−\mathrm{ln}\:\mathrm{64}\centerdot{x}} \right)\:=\:−\mathrm{ln}\:\mathrm{64} \\ $$$$−\mathrm{ln}\:\mathrm{64}\centerdot{x}\centerdot{e}^{−\mathrm{ln}\:\mathrm{64}\centerdot{x}} \:=\:−\mathrm{ln}\:\mathrm{64} \\ $$$$−\mathrm{ln}\:\mathrm{64}\centerdot{x}\:=\:{y} \\ $$$${ye}^{{y}} \:=\:−\mathrm{ln}\:\mathrm{64} \\ $$$${y}\:=\:{W}\left(−\mathrm{ln}\:\mathrm{64}\right)\:\:\:\:{W}\::\:{Lambert}\:{function} \\ $$$${x}\:=\:−\:\frac{{W}\left(−\mathrm{ln}\:\mathrm{64}\right)}{\mathrm{ln}\:\mathrm{64}} \\ $$
Commented by mrW1 last updated on 22/Apr/17
W(−ln 64) is not valid! For Lambert function W(t) t must be ≥−(1/e) ≈−0.368 but −ln 64≈−4.159 !!!
$${W}\left(−\mathrm{ln}\:\mathrm{64}\right)\:{is}\:{not}\:{valid}!\: \\ $$$$ \\ $$$${For}\:{Lambert}\:{function}\:{W}\left({t}\right) \\ $$$${t}\:{must}\:{be}\:\geqslant−\frac{\mathrm{1}}{{e}}\:\approx−\mathrm{0}.\mathrm{368} \\ $$$${but}\:−\mathrm{ln}\:\mathrm{64}\approx−\mathrm{4}.\mathrm{159}\:!!! \\ $$
Commented by tawa last updated on 23/Apr/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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