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Question Number 64601 by LPM last updated on 19/Jul/19
10^x =x^(1000) ⇒ x =?
$$\:\:\:\:\:\mathrm{10}^{{x}} ={x}^{\mathrm{1000}} \:\Rightarrow\:{x}\:=? \\ $$
Answered by mr W last updated on 19/Jul/19
x^(1000) =10^x ⇒x=±10^(x/(1000)) ⇒x=±e^((x/(1000))×ln 10) ⇒x×e^(−(x/(1000))×ln 10) =±1 ⇒(−(x/(1000))×ln 10)×e^(−(x/(1000))×ln 10) =∓((ln 10)/(1000)) ⇒−(x/(1000))×ln 10=W(∓((ln 10)/(1000))) ← Lambert W function ⇒x=−((1000)/(ln 10))×W(∓((ln 10)/(1000))) ⇒x= { ((−((1000)/(ln 10))×W(−((ln 10)/(1000)))= { ((−((1000×(−0.0023079))/(ln 10))=1.00231)),((−((1000×(−8.1747762))/(ln 10))=3550.2602)) :})),((−((1000)/(ln 10))×W(((ln 10)/(1000)))=−((1000×0.0022973)/(ln 10))=−0.99771)) :} there are totally 3 solutions: x=−0.99771, 1.00231, 3550.2602
$${x}^{\mathrm{1000}} =\mathrm{10}^{{x}} \\ $$$$\Rightarrow{x}=\pm\mathrm{10}^{\frac{{x}}{\mathrm{1000}}} \\ $$$$\Rightarrow{x}=\pm{e}^{\frac{{x}}{\mathrm{1000}}×\mathrm{ln}\:\mathrm{10}} \\ $$$$\Rightarrow{x}×{e}^{−\frac{{x}}{\mathrm{1000}}×\mathrm{ln}\:\mathrm{10}} =\pm\mathrm{1} \\ $$$$\Rightarrow\left(−\frac{{x}}{\mathrm{1000}}×\mathrm{ln}\:\mathrm{10}\right)×{e}^{−\frac{{x}}{\mathrm{1000}}×\mathrm{ln}\:\mathrm{10}} =\mp\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{1000}} \\ $$$$\Rightarrow−\frac{{x}}{\mathrm{1000}}×\mathrm{ln}\:\mathrm{10}=\mathbb{W}\left(\mp\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{1000}}\right)\:\:\leftarrow\:{Lambert}\:{W}\:{function} \\ $$$$\Rightarrow{x}=−\frac{\mathrm{1000}}{\mathrm{ln}\:\mathrm{10}}×\mathbb{W}\left(\mp\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{1000}}\right) \\ $$$$\Rightarrow{x}=\begin{cases}{−\frac{\mathrm{1000}}{\mathrm{ln}\:\mathrm{10}}×\mathbb{W}\left(−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{1000}}\right)=\begin{cases}{−\frac{\mathrm{1000}×\left(−\mathrm{0}.\mathrm{0023079}\right)}{\mathrm{ln}\:\mathrm{10}}=\mathrm{1}.\mathrm{00231}}\\{−\frac{\mathrm{1000}×\left(−\mathrm{8}.\mathrm{1747762}\right)}{\mathrm{ln}\:\mathrm{10}}=\mathrm{3550}.\mathrm{2602}}\end{cases}}\\{−\frac{\mathrm{1000}}{\mathrm{ln}\:\mathrm{10}}×\mathbb{W}\left(\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{1000}}\right)=−\frac{\mathrm{1000}×\mathrm{0}.\mathrm{0022973}}{\mathrm{ln}\:\mathrm{10}}=−\mathrm{0}.\mathrm{99771}}\end{cases} \\ $$$${there}\:{are}\:{totally}\:\mathrm{3}\:{solutions}: \\ $$$${x}=−\mathrm{0}.\mathrm{99771},\:\mathrm{1}.\mathrm{00231},\:\mathrm{3550}.\mathrm{2602} \\ $$
Commented by Tony Lin last updated on 19/Jul/19
3550?
$$\mathrm{3550}? \\ $$
Commented by Tony Lin last updated on 19/Jul/19
how do you calculate Lambert W function? what calculator can calculate this function?
$${how}\:{do}\:{you}\:{calculate}\:{Lambert}\:{W}\:{function}? \\ $$$${what}\:{calculator}\:{can}\:{calculate}\:{this}\:{function}? \\ $$
Commented by mr W last updated on 19/Jul/19
x^(1000) =10^x ⇒1000×log x=x with x=3550.2602 1000×log 3550.2602=3550.2602 i.e. x=3550.2602 is a correct solution.
$${x}^{\mathrm{1000}} =\mathrm{10}^{{x}} \\ $$$$\Rightarrow\mathrm{1000}×\mathrm{log}\:{x}={x} \\ $$$${with}\:{x}=\mathrm{3550}.\mathrm{2602} \\ $$$$\mathrm{1000}×\mathrm{log}\:\mathrm{3550}.\mathrm{2602}=\mathrm{3550}.\mathrm{2602} \\ $$$${i}.{e}.\:{x}=\mathrm{3550}.\mathrm{2602}\:{is}\:{a}\:{correct}\:{solution}. \\ $$
Commented by mr W last updated on 19/Jul/19
Lambert function W(a) is the root of equation xe^x =a, if you don′t have a calculator, you can use graphic methods to get its value. i can recomment you a very nice app called GRAPHER.
$${Lambert}\:{function}\:{W}\left({a}\right)\:{is}\:{the}\:{root} \\ $$$${of}\:{equation}\:{xe}^{{x}} ={a},\:{if}\:{you}\:{don}'{t} \\ $$$${have}\:{a}\:{calculator},\:{you}\:{can}\:{use}\:{graphic} \\ $$$${methods}\:{to}\:{get}\:{its}\:{value}. \\ $$$${i}\:{can}\:{recomment}\:{you}\:{a}\:{very}\:{nice}\:{app} \\ $$$${called}\:{GRAPHER}. \\ $$

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