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solve-x-x-x-2019-2019-




Question Number 71824 by mr W last updated on 20/Oct/19
solve   x^x^x^(2019)   =2019
$${solve}\: \\ $$$${x}^{{x}^{{x}^{\mathrm{2019}} } } =\mathrm{2019} \\ $$
Commented by mr W last updated on 21/Oct/19
we can see:  if we assume x^(2019) =2019, we will get  x^x^(2019)  =x^(2019) =2019  x^x^x^(2019)   =x^(2019) =2019  x^x^x^x^(2019)    =x^(2019) =2019  ....  i.e. the assumption x^(2019) =2019 is correct.  from x^(2019) =2019, we get  ⇒ x=((2019))^(1/(2019))
$${we}\:{can}\:{see}: \\ $$$${if}\:{we}\:{assume}\:{x}^{\mathrm{2019}} =\mathrm{2019},\:{we}\:{will}\:{get} \\ $$$${x}^{{x}^{\mathrm{2019}} } ={x}^{\mathrm{2019}} =\mathrm{2019} \\ $$$${x}^{{x}^{{x}^{\mathrm{2019}} } } ={x}^{\mathrm{2019}} =\mathrm{2019} \\ $$$${x}^{{x}^{{x}^{{x}^{\mathrm{2019}} } } } ={x}^{\mathrm{2019}} =\mathrm{2019} \\ $$$$…. \\ $$$${i}.{e}.\:{the}\:{assumption}\:{x}^{\mathrm{2019}} =\mathrm{2019}\:{is}\:{correct}. \\ $$$${from}\:{x}^{\mathrm{2019}} =\mathrm{2019},\:{we}\:{get} \\ $$$$\Rightarrow\:{x}=\sqrt[{\mathrm{2019}}]{\mathrm{2019}} \\ $$
Answered by MJS last updated on 20/Oct/19
usually x^x^x^(2019)   =x^((x^((x^(2019) )) ))   x^((x^((x^(2019) )) )) =2019 ⇒  ⇒ 1.00377648280<x<1.00377648281    but if you mean  ((x^x )^x )^(2019) =2019 ⇒  ⇒ 1.00374827787<x<1.00374827788
$$\mathrm{usually}\:{x}^{{x}^{{x}^{\mathrm{2019}} } } ={x}^{\left({x}^{\left({x}^{\mathrm{2019}} \right)} \right)} \\ $$$${x}^{\left({x}^{\left({x}^{\mathrm{2019}} \right)} \right)} =\mathrm{2019}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{1}.\mathrm{00377648280}<{x}<\mathrm{1}.\mathrm{00377648281} \\ $$$$ \\ $$$$\mathrm{but}\:\mathrm{if}\:\mathrm{you}\:\mathrm{mean} \\ $$$$\left(\left({x}^{{x}} \right)^{{x}} \right)^{\mathrm{2019}} =\mathrm{2019}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{1}.\mathrm{00374827787}<{x}<\mathrm{1}.\mathrm{00374827788} \\ $$
Commented by mr W last updated on 21/Oct/19
thanks sir!  the first interpretation is meant.  the exact value is x=((2019))^(1/(2019))
$${thanks}\:{sir}! \\ $$$${the}\:{first}\:{interpretation}\:{is}\:{meant}. \\ $$$${the}\:{exact}\:{value}\:{is}\:{x}=\sqrt[{\mathrm{2019}}]{\mathrm{2019}} \\ $$
Commented by TawaTawa last updated on 21/Oct/19
Please workings sir
$$\mathrm{Please}\:\mathrm{workings}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 22/Oct/19
see above.  the result is the same, no matter   how many times x occurs in the  equation.
$${see}\:{above}. \\ $$$${the}\:{result}\:{is}\:{the}\:{same},\:{no}\:{matter}\: \\ $$$${how}\:{many}\:{times}\:{x}\:{occurs}\:{in}\:{the} \\ $$$${equation}. \\ $$
Answered by mind is power last updated on 23/Oct/19
x^x^(2019)  ln(x)=ln(2019)  ⇒x^(2019) ln(x)+ln(ln(x))=ln(ln(2019))  ⇒x^(2019) ln(x)=ln(((ln(2019))/(ln(x))))  u=((ln(2019))/(ln(x)))⇒ln(x)=((ln(2019))/u)=⇒x=2019^(1/u)   ⇒2019^((2019)/u) .((ln(2019))/u)=ln(u)  ⇒2019^(2019^((2019)/u) ) =u^u^1  =S  1=(u/u)  ⇒S⇔2019^(2019^((2019)/u) ) =u^u^(u/u)  ⇒u=2019  ln(x)=((ln(2019))/(2019))⇒x=((2019))^(1/(2019))
$$\mathrm{x}^{\mathrm{x}^{\mathrm{2019}} } \mathrm{ln}\left(\mathrm{x}\right)=\mathrm{ln}\left(\mathrm{2019}\right) \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2019}} \mathrm{ln}\left(\mathrm{x}\right)+\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{x}\right)\right)=\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{2019}\right)\right) \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2019}} \mathrm{ln}\left(\mathrm{x}\right)=\mathrm{ln}\left(\frac{\mathrm{ln}\left(\mathrm{2019}\right)}{\mathrm{ln}\left(\mathrm{x}\right)}\right) \\ $$$$\mathrm{u}=\frac{\mathrm{ln}\left(\mathrm{2019}\right)}{\mathrm{ln}\left(\mathrm{x}\right)}\Rightarrow\mathrm{ln}\left(\mathrm{x}\right)=\frac{\mathrm{ln}\left(\mathrm{2019}\right)}{\mathrm{u}}=\Rightarrow\mathrm{x}=\mathrm{2019}^{\frac{\mathrm{1}}{\mathrm{u}}} \\ $$$$\Rightarrow\mathrm{2019}^{\frac{\mathrm{2019}}{\mathrm{u}}} .\frac{\mathrm{ln}\left(\mathrm{2019}\right)}{\mathrm{u}}=\mathrm{ln}\left(\mathrm{u}\right) \\ $$$$\Rightarrow\mathrm{2019}^{\mathrm{2019}^{\frac{\mathrm{2019}}{\mathrm{u}}} } =\mathrm{u}^{\mathrm{u}^{\mathrm{1}} } =\mathrm{S} \\ $$$$\mathrm{1}=\frac{\mathrm{u}}{\mathrm{u}} \\ $$$$\Rightarrow\mathrm{S}\Leftrightarrow\mathrm{2019}^{\mathrm{2019}^{\frac{\mathrm{2019}}{\mathrm{u}}} } =\mathrm{u}^{\mathrm{u}^{\frac{\mathrm{u}}{\mathrm{u}}} } \Rightarrow\mathrm{u}=\mathrm{2019} \\ $$$$\mathrm{ln}\left(\mathrm{x}\right)=\frac{\mathrm{ln}\left(\mathrm{2019}\right)}{\mathrm{2019}}\Rightarrow\mathrm{x}=\sqrt[{\mathrm{2019}}]{\mathrm{2019}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 23/Oct/19
thanks alot sir!
$${thanks}\:{alot}\:{sir}! \\ $$
Commented by mind is power last updated on 23/Oct/19
y ′re welcom
$$\mathrm{y}\:'\mathrm{re}\:\mathrm{welcom}\:\: \\ $$

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