A Nice Log Equation With Natural Log

[@FisicTrapella]

1st -> If you ln instead of log you get the same without changing of base 🙃

[@Montadhro3o_l]

In 1st method you dont have to use log
U just use ln to get it faster otherwise its both true

[@marzipanhoplite17]

Both methods are nice and easily understood.As for your compatriot I really cannot understand his negative comments about you, provided of course that the translation from Google is reliable

[@pablocopello3592]

Or, you can put: x=10^+-sqrt(log(e)).
That is, interchange e and 10 everywhere, because it can be done in the original eq.

[@md.zahirulhoque8452]

2nd was much better but you have to remember that you can solve this like that otherwise the 1st one will pop up in exm 1st 😆

[@Don-Ensley]

problem
xˡⁿ ˣ = 10
Note especially for this problem that
√[ ln (10) ] ≠ ln (√10)
Take natural logs.
[ ln (x)] ² = ln (10)
ln (x) = ± √[ ln (10) ]
There are 2 solutions.
x = e^√[ ln (10) ]
x = 1 / e^√[ ln (10) ]
answer
x ∈ { 1 / e^√[ ln (10) ], e^√[ ln (10) ] }

[@SweetSorrow777]

Log x = ln x / ln 10 That's the key.

[@misterj.a91]

Both methods seem like just making your life harder unnecessarily. Just ln both sides and then square root and you're done

[@quark67000]

x^{\ln x} = 10, not x^{lnx} = 10. Thanks.

[@barakathaider6333]

👍

[@saltukbugra-oo8fc]

abi merbaha türk müsün konuşmadan türk gibisin gibi me geldi türksen bir türk ün böyle global bir matematik kanalı açması gurur verici değilsende sıkıntı yok insanlar bir şeyler öğreniyor buda güzel bi şey kolay gelsin

[@prollysine]

logx/loge * logx = 1 , (logx)^2=loge , x=10^(V(loge)) , test , x^lnx=10 , OK ,

[@russellsharpe288]

x ^ ln x = 10 => (ln x)² = (ln x).(ln x) = ln (x^(ln x)) = ln 10 so ln x = +- sqrt(ln 10) and x = exp(+-sqrt(ln 10).

[@pspprabhat]

Lnx.lnx=ln10
Lnx=+_ _/ln10
=>×=e^_/2.303=4.567 3:21