A Mixed Equation

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[@tunneloflight]

also x = ln(1/log(e))

[@il_solito_anonimo9164]

Well it's the same thing, because 1/log(e)->1/(In(e)/In(10))->1/(1/In(10))->In(10), so In(1/log(e))=In(In(10))

[@SidneiMV]

(1/logx)lnx = ln10
eˣ= ln10
*x = ln(ln10)*

[@Khashayarissi-ob4yj]

With luck and more power to you.
hoping for more videos

[@tejpalsingh366]

e^e^x= 10

[@Gordy-io8sb]

e^0=ln0/log0
e^0=1/1
e^0=1
x=0
I don't know, I'm bad with logarithms.
E: I wrote this comment while running on like 4 or 5 hours of sleep -- I realize my mistake now.

[@il_solito_anonimo9164]

In0 and log0 are both equals to -infinity, so they don't exist in real, and complex, solutions. I think you changed the 0 and the 1, in fact log(1)=0

[@Gordy-io8sb]

@@il_solito_anonimo9164 Wait, wouldn't -∞ * (-∞)^-1 just be 1?

[@Monero_Monello]

​@@Gordy-io8sb some infinities are bigger than others, you can't simplify them like that

[@Gordy-io8sb]

@@Monero_Monello That's an elementary aspect of set theory. This is analysis. Also, you're not even using the statement correctly.

[@Monero_Monello]

​@@Gordy-io8sb whatever gets the point across is enough in my books

[@Foamea45]

e^x=lnx/(lnx/ln100 e^x=ln10=>x=ln(ln10)

[@phill3986]

😊🎉😊👍👍👍😊🎉😊

[@barberickarc3460]

If you use change of base on lnx to logx/log(e) you see the x terms cancel and you have a constant on the right side, so you're fine to ln both sides and get x = -ln(log e) ≈ 0.834

[@TheOldeCrowe]

eˣ = lnx/logx = lnx/(lnx/ln10) = ln10
x = ln(ln10)