Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com