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)\\(1)/(4x)=e^(-x \ln 2)\\-(\ln 2)/(4)=-x\ln 2e^(-x \ln 2)\\-x\ln 2=W\left(-(\ln 2)/(4)\right)\\x=-(W\left(-(\ln 2)/(4)\right))/(\ln 2)`

1.

`x=\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+√(5\ldots)}}}}}\qquad\left( x\geq0\right)\\\\x=\sqrt{13+√(5+x)}\\\\x^2=13+√(5+x)\\√(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 √(13) \vee x\leq-√(13)\wedge x\geq0 \\D:x\in[√(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+√(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.