Solving x^{ln(x)}=ex

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Nicely made my friend:) Thanks so much. May God bless you every moment of your life and i wish you a long and happy one to. Love and praises from YoonHo, South Korea, and congratulations on 1 milloin views!!! I am a big fan so please pin me!!!

SuperYoonHo

Feels good to be able to calculate it in my head

lassikokkonen

Given:
x↑(ln(x)) = ex
To find:
x

Taking natural log on both sides:
ln(x↑(ln(x))) = ln(ex)

Using ln(a↑b) = bln(a) and ln(ab) = ln(a) + ln(b):
(ln(x))·(ln(x)) = ln(e) + ln(x)

Using ln(e) = 1:
(ln(x))² = ln(x) + 1

Moving all terms to LHS:
(ln(x))² – ln(x) – 1 = 0

This is a standard quadratic equation, with two roots: φ and it's conjugate, ̅φ. So:
ln(x) = φ or ̅φ

Taking the exponent of both sides:
x = e↑φ or e↑̅φ

GirishManjunathMusic

Another way a little different : log both sides with base x :
lnx = log_x e + 1.
With cob formula, we find that log_x e = 1/lnx.
Multiply both sides by lnx and got the same equation

damiennortier

Wow. I basically took the 2nd approach(without setting t though) but and almost thought I got it wrong when I got the goden ratio.But in the end I forgot about the plus-minus sign and instead just yielded e^'golden ratio'

Jenghis-Khn

Nice problem for those who still love their "ex"
😂😂😂

debtanaysarkar

Solved by the first method. Thanks for the second method 🙂

ubtfsvx

can you teach natural logarithms from start to finish...

rustam

3rd method?
x^(lnx)=ex
Divide both members to x
(x^(lnx))/x=ex/x
x^[(lnx)-1]=e
Rescribe "e" as x^(ln in base x of e)

x^[(lnx)-1]=x^(ln in base x of e)

(lnx)-1=ln in base x of e
Change base to (ln in base x of e) in (lne)/(lnx)
(lnx)-1=(lne)/(lnx)
(lnx)^2 -lnx -1=0

Alternatively:
x^(lnx)=ex
x [x^((lnx)-1) -e]=0
x^((lnx)-1) -e=0
Change base to the e and....

drdiegocolombo

Too easy Syber I looked at it and already knew it was a golden equation

moeberry

*@ SyberMath --* There is at least one place in the video where your handwriting is sloppy/careless
to the extent that [in blue letters] t = (1 - sqrt(5))/2 looks closer to t = 1 - sqrt(5)/2.
You need to make sure the fraction bar extends sufficiently underneath a numerator.

robertveith

Как всегда великолепно. Большое спасибо

iqtgzpr

Tsk tsk. When you say that e^(t^2) = e^(t + 1) implies t^2 = t + 1 you are really taking the ln of both sides😉, so the two methods are essentially the same.

Dan-cwxu

I have started to smell the 100K subscribers

YahiaNebti

Where is your accent from? It's so sweet and cute and I love it

shlomi

Bro how to solve (ln (x²) /x) +x =0
I'm losing my mind

reaper-mf

Yes, Now i have started to smell the golden ratio too!

Jha-s-kitchen

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