@tunneloflight Ай бұрын
also x = ln(1/log(e))
@il_solito_anonimo9164 Ай бұрын
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 Ай бұрын
(1/logx)lnx = ln10 eˣ= ln10 *x = ln(ln10)*
@Khashayarissi-ob4yj Ай бұрын
With luck and more power to you. hoping for more videos
@Gordy-io8sb Ай бұрын
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 Ай бұрын
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 Ай бұрын
@@il_solito_anonimo9164 Wait, wouldn't -∞ * (-∞)^-1 just be 1?
@Monero_Monello Ай бұрын
@@Gordy-io8sb some infinities are bigger than others, you can't simplify them like that
@Gordy-io8sb Ай бұрын
@@Monero_Monello That's an elementary aspect of set theory. This is analysis. Also, you're not even using the statement correctly.
@Monero_Monello Ай бұрын
@@Gordy-io8sb whatever gets the point across is enough in my books
@tejpalsingh366 Ай бұрын
e^e^x= 10
@phill3986 Ай бұрын
😊🎉😊👍👍👍😊🎉😊
@barberickarc3460 Ай бұрын
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
@Foamea45 Ай бұрын
e^x=lnx/(lnx/ln100 e^x=ln10=>x=ln(ln10)
@TheOldeCrowe Ай бұрын
eˣ = lnx/logx = lnx/(lnx/ln10) = ln10 x = ln(ln10)
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