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Question Number 58832 by otchereabdullai@gmail.com last updated on 30/Apr/19
find x if x^2 =16^x
$$\mathrm{find}\:\mathrm{x}\:\mathrm{if}\:\mathrm{x}^{\mathrm{2}} =\mathrm{16}^{\mathrm{x}} \\ $$
Commented by tanmay last updated on 01/May/19
f(x)=x^2 −16^x f(0)<0 f(1)<0 so no root between (0,1) f(−1)=1−(1/(16))>0 so root lies between (−1,0)←location of root f(x)=x^2 −(4^x )^2 =(x+4^x )(x−4^x ) on trial if we put x=−0.5 f(−0.5)=(−0.5+4^(−0.5) )(−0.5−4^(−0.5) ) ={(−0.5)^2 −(4^(−0.5) )^2 } ={0.25−(4^(−1) )} =0.25−0.25=0 so x=−0.5 is the solution i have located the location of root and found by trial method...There must be some direct method to find the root.
$${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{16}^{{x}} \\ $$$${f}\left(\mathrm{0}\right)<\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)<\mathrm{0} \\ $$$${so}\:{no}\:{root}\:{between}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$${f}\left(−\mathrm{1}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}>\mathrm{0} \\ $$$${so}\:{root}\:{lies}\:{between}\:\left(−\mathrm{1},\mathrm{0}\right)\leftarrow{location}\:{of}\:{root} \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} −\left(\mathrm{4}^{{x}} \right)^{\mathrm{2}} =\left({x}+\mathrm{4}^{{x}} \right)\left({x}−\mathrm{4}^{{x}} \right) \\ $$$${on}\:{trial}\:{if}\:{we}\:{put}\:{x}=−\mathrm{0}.\mathrm{5} \\ $$$${f}\left(−\mathrm{0}.\mathrm{5}\right)=\left(−\mathrm{0}.\mathrm{5}+\mathrm{4}^{−\mathrm{0}.\mathrm{5}} \right)\left(−\mathrm{0}.\mathrm{5}−\mathrm{4}^{−\mathrm{0}.\mathrm{5}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left\{\left(−\mathrm{0}.\mathrm{5}\right)^{\mathrm{2}} −\left(\mathrm{4}^{−\mathrm{0}.\mathrm{5}} \right)^{\mathrm{2}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left\{\mathrm{0}.\mathrm{25}−\left(\mathrm{4}^{−\mathrm{1}} \right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}.\mathrm{25}−\mathrm{0}.\mathrm{25}=\mathrm{0} \\ $$$${so}\:{x}=−\mathrm{0}.\mathrm{5}\:{is}\:{the}\:{solution} \\ $$$${i}\:{have}\:{located}\:{the}\:{location}\:{of}\:{root}\:{and}\:{found} \\ $$$${by}\:{trial}\:{method}…\boldsymbol{{T}}{here}\:{must}\:{be}\:{some}\:{direct} \\ $$$${method}\:{to}\:{find}\:{the}\:{root}. \\ $$$$ \\ $$
Commented by otchereabdullai@gmail.com last updated on 30/Apr/19
thanks alot prof tanmay
$$\mathrm{thanks}\:\mathrm{alot}\:\mathrm{prof}\:\mathrm{tanmay} \\ $$
Commented by George Mark Samuel last updated on 30/Apr/19
x^2 =16^x x^2 =4^((2)x) x^2 =2^(2(2x)) x^2 =2^(4x) x=2 .∣2=4x ∣divide both sides by 4 ∣(2/4)=((4x)/4) ∣x=(1/2) ∴x=2 or x=(1/(2.))
$$\mathrm{x}^{\mathrm{2}} =\mathrm{16}^{\mathrm{x}} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{4}^{\left(\mathrm{2}\right)\mathrm{x}} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}\left(\mathrm{2x}\right)} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{2}^{\mathrm{4x}} \\ $$$$\mathrm{x}=\mathrm{2}\:.\mid\mathrm{2}=\mathrm{4x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{divide}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{by}\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\frac{\mathrm{2}}{\mathrm{4}}=\frac{\mathrm{4x}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\therefore\mathrm{x}=\mathrm{2}\:\mathrm{or}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}.} \\ $$$$ \\ $$
Commented by MJS last updated on 01/May/19
it′s wrong x=2 ⇒ x^2 =4 ∧ 16^x =256 but 4≠256 x=(1/2) ⇒ x^2 =(1/4) ∧ 16^x =4 but (1/4)≠4
$$\mathrm{it}'\mathrm{s}\:\mathrm{wrong} \\ $$$${x}=\mathrm{2}\:\Rightarrow\:{x}^{\mathrm{2}} =\mathrm{4}\:\wedge\:\mathrm{16}^{{x}} =\mathrm{256}\:\mathrm{but}\:\mathrm{4}\neq\mathrm{256} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\:\wedge\:\mathrm{16}^{{x}} =\mathrm{4}\:\mathrm{but}\:\frac{\mathrm{1}}{\mathrm{4}}\neq\mathrm{4} \\ $$
Answered by MJS last updated on 30/Apr/19
obviously x≥0 ⇒ 16^x >x^2 x≤−1 ⇒ x^2 >16^x ⇒ −1<x<0 trying we find x=−(1/2) ⇒ (−(1/2))^2 =16^(−(1/2)) =(1/4)
$$\mathrm{obviously} \\ $$$${x}\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{16}^{{x}} >{x}^{\mathrm{2}} \\ $$$${x}\leqslant−\mathrm{1}\:\Rightarrow\:{x}^{\mathrm{2}} >\mathrm{16}^{{x}} \\ $$$$\Rightarrow\:−\mathrm{1}<{x}<\mathrm{0} \\ $$$$\mathrm{trying}\:\mathrm{we}\:\mathrm{find} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{16}^{−\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by otchereabdullai@gmail.com last updated on 30/Apr/19
thanks prof
$$\mathrm{thanks}\:\mathrm{prof} \\ $$
Answered by mr W last updated on 01/May/19
direct method using Lambert function: x^2 =16^x ⇒x=±16^(x/2) ⇒x=±e^((x ln 16)/2) ⇒xe^(−((x ln 16)/2)) =±1 ⇒(−((x ln 16)/2))e^(−((x ln 16)/2)) =∓((ln 16)/2) ⇒−((x ln 16)/2)=W(∓((ln 16)/2)) ⇒x=−(2/(ln 16))×W(∓((ln 16)/2)) ←no real solution for W(−((ln 16)/2))! ⇒x=−((2×0.69314718)/(ln 16))=−0.5 Bonus: in general, if x^2 =a^x ⇒x=−(2/(ln a))×W(±((ln a)/2)) e.g. a=2, i.e. x^2 =2^x ⇒x=−(2/(ln 2))×W(±((ln 2)/2))= { ((−(2/(ln 2))×W(((ln 2)/2))=−(2/(ln 2))×0.26570574=−0.766665)),((−(2/(ln 2))×W(−((ln 2)/2))= { ((−(2/(ln 2))×(−0.69314718)=2)),((−(2/(ln 2))×(−1.38629436)=4)) :})) :}
$${direct}\:{method}\:{using}\:{Lambert}\:{function}: \\ $$$${x}^{\mathrm{2}} =\mathrm{16}^{{x}} \\ $$$$\Rightarrow{x}=\pm\mathrm{16}^{\frac{{x}}{\mathrm{2}}} \\ $$$$\Rightarrow{x}=\pm{e}^{\frac{{x}\:\mathrm{ln}\:\mathrm{16}}{\mathrm{2}}} \\ $$$$\Rightarrow{xe}^{−\frac{{x}\:\mathrm{ln}\:\mathrm{16}}{\mathrm{2}}} =\pm\mathrm{1} \\ $$$$\Rightarrow\left(−\frac{{x}\:\mathrm{ln}\:\mathrm{16}}{\mathrm{2}}\right){e}^{−\frac{{x}\:\mathrm{ln}\:\mathrm{16}}{\mathrm{2}}} =\mp\frac{\mathrm{ln}\:\mathrm{16}}{\mathrm{2}} \\ $$$$\Rightarrow−\frac{{x}\:\mathrm{ln}\:\mathrm{16}}{\mathrm{2}}=\mathbb{W}\left(\mp\frac{\mathrm{ln}\:\mathrm{16}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}=−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{16}}×\mathbb{W}\left(\mp\frac{\mathrm{ln}\:\mathrm{16}}{\mathrm{2}}\right)\:\leftarrow{no}\:{real}\:{solution}\:{for}\:{W}\left(−\frac{\mathrm{ln}\:\mathrm{16}}{\mathrm{2}}\right)! \\ $$$$\Rightarrow{x}=−\frac{\mathrm{2}×\mathrm{0}.\mathrm{69314718}}{\mathrm{ln}\:\mathrm{16}}=−\mathrm{0}.\mathrm{5} \\ $$$$ \\ $$$${Bonus}: \\ $$$${in}\:{general},\:{if}\:{x}^{\mathrm{2}} ={a}^{{x}} \\ $$$$\Rightarrow{x}=−\frac{\mathrm{2}}{\mathrm{ln}\:{a}}×\mathbb{W}\left(\pm\frac{\mathrm{ln}\:{a}}{\mathrm{2}}\right) \\ $$$${e}.{g}.\:{a}=\mathrm{2},\:{i}.{e}.\:{x}^{\mathrm{2}} =\mathrm{2}^{{x}} \\ $$$$\Rightarrow{x}=−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}×\mathbb{W}\left(\pm\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)=\begin{cases}{−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}×{W}\left(\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)=−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}×\mathrm{0}.\mathrm{26570574}=−\mathrm{0}.\mathrm{766665}}\\{−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}×{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)=\begin{cases}{−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}×\left(−\mathrm{0}.\mathrm{69314718}\right)=\mathrm{2}}\\{−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}×\left(−\mathrm{1}.\mathrm{38629436}\right)=\mathrm{4}}\end{cases}}\end{cases} \\ $$
Commented by otchereabdullai@gmail.com last updated on 01/May/19
thank you soo much prof W
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{soo}\:\mathrm{much}\:\mathrm{prof}\:\mathrm{W} \\ $$
Commented by malwaan last updated on 02/May/19
How can I calc. W function?
$${How}\:{can}\:{I}\:{calc}.\:\mathbb{W}\:{function}? \\ $$
Commented by mr W last updated on 02/May/19
if you can′t find a proper calculator, you can get the function value in this way: the root(s) of eqn. x e^x =a is (are) the function value(s) of W(a).
$${if}\:{you}\:{can}'{t}\:{find}\:{a}\:{proper}\:{calculator}, \\ $$$${you}\:{can}\:{get}\:{the}\:{function}\:{value}\:{in} \\ $$$${this}\:{way}: \\ $$$${the}\:{root}\left({s}\right)\:{of}\:{eqn}. \\ $$$${x}\:{e}^{{x}} ={a} \\ $$$${is}\:\left({are}\right)\:{the}\:{function}\:{value}\left({s}\right)\:{of}\:\mathbb{W}\left({a}\right). \\ $$
Commented by malwaan last updated on 03/May/19
I will try thank you sir
$${I}\:{will}\:{try}\: \\ $$$${thank}\:{you}\:{sir} \\ $$

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