Question

Find 2^x=4x, find x ​

Answer 1

Whatever... I'll try to solve the first two problems.

$2^x=4x\\4x=e^{\ln 2^x}\\4x=e^{x\ln 2}\\\dfrac{1}{4x}=e^{-x \ln 2}\\-\dfrac{\ln 2}{4}=-x\ln 2e^{-x \ln 2}\\-x\ln 2=W\left(-\dfrac{\ln 2}{4}\right)\\x=-\dfrac{W\left(-\dfrac{\ln 2}{4}\right)}{\ln 2}$

1.

$x=\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5\ldots}}}}}}\qquad\left( x\geq0\right)\\\\x=\sqrt{13+\sqrt{5+x}}\\\\x^2=13+\sqrt{5+x}\\\sqrt{5+x}=x^2-13\\\\D:5+x\geq0 \wedge x^2-13\geq 0\wedge x\geq 0\\D:x\geq-5 \wedge x^2\geq13\wedge x\geq 0\\D:x\geq \sqrt{13} \vee x\leq-\sqrt{13}\wedge x\geq0 \\D:x\in[\sqrt{13},\infty)\\\\ 5+x=x^4-26x^2+169\\x^4-26x^2-x+164=0\\\vdots\\(x - 4) (x^3 + 4 x^2 - 10 x - 41) = 0\\x-4=0\\x=4$

One solution is $x=4$, and $4\in D$, therefore$\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5\ldots}}}}}}=4$.

We can ignore other possible solutions of the polynomial equation, because the expression can not be equal to two different values simultaneously.