Solving a Logarithmic Exponential Equation

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[@angelmendez-rivera351]

x^(1/log(x)) = 10, as tends to be the case with these things. So, 10·log(x) = 1. The rest is trivial. In general, for the logarithm in base b, log(b, x), we have that x^(1/log(b, x)) = b.

[@owlsmath]

Wow! That really simplified nicely. Easier than i was expecting.

[@Skank_and_Gutterboy]

Yes!

[@Skank_and_Gutterboy]

Holy smokes, pretty sure this is the first time that I've worked the problem just like it was done in the video. Very enjoyable problem!

[@SyberMath]

Glad to hear that!

[@minhdoantuan8807]

The condition of x is x >0 and x 1
Using 1/log(x) = log_x(10), switching the bases, the equation becomes:
10 log(x) =1, log(x) = 1/10, x = 10^(1/10) is the only solution.

[@erpaninozzo]

it would be much simpler say that x^(1/log(x)) = 10^(log(x)/log(x)) = 10 so 10*log(x) = 1 --> log(x) = 1/10 --> x = 10^(1/10)

[@user-lh5hl4sv8z]

why

[@CriticSimon]

I don’t think so

[@angelmendez-rivera351]

@@CriticSimon I agree, it would be simpler. It is just straightforward simplification.

[@moeberry8226]

Exactly, that’s way simpler.

[@CriticSimon]

@@angelmendez-rivera351 Sounds good!

[@manuelgonzales2570]

Fantastic!!

[@jamesyu1093]

subbing y = log x and solving 10 y = 1 for y makes the problem trivial

[@lavoiedereussite922]

I like your demonstration.

[@Trip1of3]

Nice, I love a good log problem. ♥️

[@SyberMath]

Thanks!

[@sk8erJG95]

Very fun! You could also go the change of variable route by letting u = logx, so x = 10^u, and the LHS of the equation becomes u*(10^u)^(1/u) = 10u.
Then 10u = 1 when u = 1/10, so logx = 1/10 and x = 10^(1/10)

[@JaseewaJasee]

you always manage to exceed my expectations with your content!

[@SyberMath]

Happy to hear that! 😍

[@allanmarder456]

log(x) isn't 0 because 1/log(x) would be undefined. Let a=1/log(x). Then (x^a)( log(x))=1 dividing by log(x) we get x^a = a. This is satisfied if x= a^(1/a).
Taking logs of both sides log(x) = (1/a)(log(a), Multiply by a gives... a times log(x) =log(a). Since a=1/log(x) we have 1=log(a) or 10^1 = a that is a=10.
So x=a^(1/a) and x=10^(1/10)

[@moeberry8226]

Another Masterpiece

[@SyberMath]

Thank you! 💖

[@sebastianviacava743]

On a video of your channel i guess you have analyzed two cases pair or unpair in same equation can you send me the link thanks!

[@scottleung9587]

Got it!

[@ckdlinked]

Thank you for your genius problems, when I heard the sound ‘1st, 2nd, 3rd’, Oh my god, I was lucky I was not with you in high school! Haha !!!

[@SyberMath]

You are welcome! 😁

[@ulrichkaiser3794]

The blue graph is y = 10 * log(x)
;-)
cheers, Uli

[@pranavamali05]

Thnku

[@malchiktamagochi8268]

in what application u r solving that equation?

[@SyberMath]

Notability

[@ProfessorJoeyWu]

it got me wrong when I converted X^(1/logX) to X^(-logX).
I was thinking it originally was 1/X^(logX)...

[@devondevon4366]

Answer x=10^1/10 or 1.2589 or 10^0.1
x^1/logx. logx =1
x^ log x 10 = 1/logx change 1/log 10 x to log x 10, then divide both sides by logx
x^log x 10 = logx 10
10 =log x 10 (n^log n p =p)
x^10= 10 ( change to exponential form)
x = 10 ^ 1/10 raised both sides to the power of 1/10 Answer
x= 1.258925 or 10^1/10 answer

[@SuperYoonHo]

thanks so much can you please do functional equtions or number theory

[@SyberMath]

Np. Sure!

[@IbnEssaChannel]

Use logx = y
and solve it

[@devondevon4366]

There are a number of ways to do it.
x^1/logx = 1/logx (bring log x to the right)
x^logx 10= log x 10 rules of log
10 = log x 10 rules of log
x^10=10
x= 10^1/10 answer

[@redareda868]

4^x+3^2y=6Can you explain the solution to this issue

[@JSSTyger]

1/e seems to be the only answer...presuming log(x) means ln(x)

[@damiennortier8942]

1 + log(logx) = 0
Log x = 0,1
X = 10^0,1

[@BabisKoutroulis]

Brilliant solution!

[@damiennortier8942]

@@BabisKoutroulis thanks

[@brinzanalexandru2150]

It would be much easier to set x=10^n

[@gabcalvert5856]

10^1/10=1.Log1=0. Therefore we conclude that the equation is false!

[@rahulpaul5539]

Take log(x)=t and x=10^t.........then it's looks like more easy

[@ChefSalad]

You totally missed a trick with this one. That trick comes from logₕx=logₐx/logₐb, and thus 1/logₕx=logₐb/logₐx. If we let a=x, then we get the trick: 1/logₕx=logₓb. This means x^(1/logₕx)=x^logₓb=b. Thus, our equation goes like 1=x^(1/logₕx)*logₕx=b*logₕx ⇒ logₕx=1/b ⇒ x=b^(1/b). If the base b=10, then x=10^(1/10), which is your answer.

[@giuseppemalaguti435]

x=(10)^1/10

[@indrajityadav9140]

If I take the base is e

[@SyberMath]

log is base 10

[@AxiePlays601]

10^(1\10)???

[@indrajityadav9140]

My answer is e only

[@euleri0]

10^(1/10)