Finding The Inverse of xln(x)

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Опубликовано:

18 сен 2024

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Комментарии : 11

@bhavydugar6665 4 дня назад

😭that is literally one of the definitions of the lambert log

@scottleung9587 4 дня назад

Cool!

@neuralwarp 3 дня назад

And Hellevrivaan to you too!

@nopegaming2029 4 дня назад

you can just assume u= ln(y) , y= e^u , and solve for x when finding the inverse so W(x) = u W(x) = ln(y) e^W(x) = y

@phill3986 4 дня назад

😀👍✌️✌️👍😀

@Chrisoikmath_ 4 дня назад

Accept for all x>1.

@DJ.Nihad.elizade596 4 дня назад

Please find the inverse of cubic polynomial Ax³ +bx² +cx +d and exponential function x^x

@Mediterranean81 4 дня назад

Let y= x^x ln (y) = x ln (x) ln (y) = ln (x) e^ln (x) W(ln(y))= ln (x) e^W(ln y) = x So f (x) = x^x f^-1(x) = e^W(ln (x))

@SyberMath 4 дня назад

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-bk0uWApJQk0.html

@Don-Ensley 2 дня назад

problem f(x) = x ln x ≡ y f⁻¹(x) = ? replace x with y and y with x then find y. x = y ln y Solve for y. Move the exponent back up. x = ln y ʸ replace y with the equivalent eˡⁿʸ x = ln ( eˡⁿʸ ^ eˡⁿʸ ) = eˡⁿʸ ln ( eˡⁿʸ ) = ln y eˡⁿʸ This is a Lambert's W form. ln y = W(x) y = eᵂ⁽ˣ⁾ answer f⁻¹(x) = eᵂ⁽ˣ⁾

@SyberMath 8 часов назад

As always, nice!