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