y=lnx and y=kx are tangent

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[@SyberMath]

Hello everyone, Starting Wednesday, December 1st, (in 2 days), the videos will PREMIERE 1 hour and 15 minutes earlier…😉🙃😍

[@abdulllllahhh]

YAY

[@musicsubicandcebu1774]

In general - for any log base a, x is always e, and k = log of e (base a) divided by e. A solution can also be found using the exponential form . . . a^x = kx.

[@miguelcerna7406]

I remembered from your previous video over 22^21 or 21^22 you used lnx/x , we found max at e,1/e. ==> k = 1/e. Not rigorous but im ok with that. I also think that 1/e being the magical slope constant that makes y=x tangent to y=lnx just simply elegant. Nice video.

[@bomboid]

Finally one I actually got easily. Don't know if my method was technically correct tho: d/dx(lnx)=d/dx(kx) k=1/x lnx=1 k=1/e

[@leif1075]

That's exsctly what I did! Ues because ar the point they meet the slope of the tangent of ln x equals the slope of the line kx so our method is valid.

[@tambuwalmathsclass]

Great video

[@SyberMath]

Thank you! 😊

[@broytingaravsol]

k=1/x, x=1/k, ln(1/k)=1, k=1/e

[@miro.s]

It can have nice application in inequalities and in linearization the problem (moving difficult nonlinear problem into linear algebra problem) where we replace ln x by its upper boud x/e

[@SyberMath]

Interesting!

[@ahmadmazbouh]

4:40 i typed the comment asking about this already and i was a (?) away from posting it

[@7styson123]

Super

[@SyberMath]

Thanks!

[@jarikosonen4079]

Then just change y=kx into y=kx+c to get more solutions...

[@snejpu2508]

Nice little problem. : )

[@SyberMath]

Thanks!

[@studyforyou6794]

Best of the best 😘😘😘

[@SyberMath]

Thank you for the kind words! 😊💖

[@7styson123]

I love maths

[@SyberMath]

Great to hear!

[@CadeiraGritante]

arctan(2)

[@SyberMath]

Why?

[@jmart474]

k=y' y=lnx; y=kx y'=1/x so this happens at y=1 lnx=1 and x=e; k=1/e Funny.👍

[@SyberMath]

It is

[@giuseppemalaguti435]

x=W(-k) /(-k) =1/k.(coefficiente angolare uguale) ....W(-k)=-1

[@ether794]

Non credevo di trovare un italiano qui

[@ieatgarbage8771]

lnx=kx 1/x=k x=1/k ln1/k=1 1/k=e k=1/e