Question Number 69082 by Henri Boucatchou last updated on 19/Sep/19
![Solve x^2 = 16^x](https://www.tinkutara.com/question/Q69082.png)
$$\boldsymbol{{Solve}}\:\:\boldsymbol{{x}}^{\mathrm{2}} \:=\:\mathrm{16}^{\boldsymbol{{x}}} \\ $$
Commented by MJS last updated on 19/Sep/19
![f(x)=16^x −x^2 f(−1)=−((15)/(16)) f(0)=1 f(−(1/2))=0 f′(x)=16^x 4ln 2 −2x >0 ∀x∈R ⇒ ⇒ no other real solution](https://www.tinkutara.com/question/Q69086.png)
$${f}\left({x}\right)=\mathrm{16}^{{x}} −{x}^{\mathrm{2}} \\ $$$$ \\ $$$${f}\left(−\mathrm{1}\right)=−\frac{\mathrm{15}}{\mathrm{16}} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$ \\ $$$${f}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$ \\ $$$${f}'\left({x}\right)=\mathrm{16}^{{x}} \mathrm{4ln}\:\mathrm{2}\:−\mathrm{2}{x}\:>\mathrm{0}\:\forall{x}\in\mathbb{R}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{other}\:\mathrm{real}\:\mathrm{solution} \\ $$
Answered by mr W last updated on 19/Sep/19
![x^2 =16^x =2^(4x) x=±2^(2x) x=±e^(2xln 2) xe^(−2xln 2) =±1 (−2xln 2)e^(−2xln 2) =±2 ln 2 ⇒−2xln 2=W(±2ln 2) ⇒x=−((W(±2ln 2))/(2 ln 2)) real solution: ⇒x=−((W(2ln 2))/(2 ln 2))=−((0.693147)/(2ln 2))=−(1/2)](https://www.tinkutara.com/question/Q69083.png)
$${x}^{\mathrm{2}} =\mathrm{16}^{{x}} =\mathrm{2}^{\mathrm{4}{x}} \\ $$$${x}=\pm\mathrm{2}^{\mathrm{2}{x}} \\ $$$${x}=\pm{e}^{\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}} \\ $$$${xe}^{−\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}} =\pm\mathrm{1} \\ $$$$\left(−\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}\right){e}^{−\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}} =\pm\mathrm{2}\:\mathrm{ln}\:\mathrm{2} \\ $$$$\Rightarrow−\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}={W}\left(\pm\mathrm{2ln}\:\mathrm{2}\right) \\ $$$$\Rightarrow{x}=−\frac{{W}\left(\pm\mathrm{2ln}\:\mathrm{2}\right)}{\mathrm{2}\:\mathrm{ln}\:\mathrm{2}} \\ $$$${real}\:{solution}: \\ $$$$\Rightarrow{x}=−\frac{{W}\left(\mathrm{2ln}\:\mathrm{2}\right)}{\mathrm{2}\:\mathrm{ln}\:\mathrm{2}}=−\frac{\mathrm{0}.\mathrm{693147}}{\mathrm{2ln}\:\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$