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at 0:37 you assumed x≥0 in which there is no solution, but there is for x
If you just apply log on both sides you will end up with e^x/2 = x isn't that better simplified value?
how is that even a solution? x is on both sides, that's not a value. All you did was write the expression in a different way.
@pandee6923 my bad I am stupid
Thanks for the video. Are you saying that the Lambert function does not yield a real solution but there may still be one? I plugged y=x^2 and y=e^x into desmos and found an approximate solution
Orx^2 = e^xx^(1/x) = e^(1/2)1/x * ln x= 1/2e^ln(1/x) * ln x = 1/2-lnx * e^ln(1/x) = -1/2ln(1/x) * e^ln(1/x) = -1/2ln(1/x) = W(-1/2)ln x = -W(-1/2)x = e^[-W(-1/2)]
Why can't you solve using the basic logarithmic function?
you won't get any sort of linear equation you can't solve it in another way than the video
@iskenderyahiaoui1844 x = 2 log x if we apply log on both sides
@@greatyeet2988 yes okay and then ?You just wrote the équation differently it won't help at all u'll always endup using the w fonction
@@iskenderyahiaoui1844 then Diffrenciate on both sides
@@greatyeet2988 in the case of an équation x isn't a variable it's a constant By differentiating u'll get 0=0
answer is about -0.703467
at 0:37 you assumed x≥0 in which there is no solution, but there is for x
If you just apply log on both sides you will end up with e^x/2 = x isn't that better simplified value?
how is that even a solution? x is on both sides, that's not a value. All you did was write the expression in a different way.
@pandee6923 my bad I am stupid
Thanks for the video. Are you saying that the Lambert function does not yield a real solution but there may still be one? I plugged y=x^2 and y=e^x into desmos and found an approximate solution
Or
x^2 = e^x
x^(1/x) = e^(1/2)
1/x * ln x= 1/2
e^ln(1/x) * ln x = 1/2
-lnx * e^ln(1/x) = -1/2
ln(1/x) * e^ln(1/x) = -1/2
ln(1/x) = W(-1/2)
ln x = -W(-1/2)
x = e^[-W(-1/2)]
Why can't you solve using the basic logarithmic function?
you won't get any sort of linear equation you can't solve it in another way than the video
@iskenderyahiaoui1844 x = 2 log x if we apply log on both sides
@@greatyeet2988 yes okay and then ?
You just wrote the équation differently it won't help at all u'll always endup using the w fonction
@@iskenderyahiaoui1844 then Diffrenciate on both sides
@@greatyeet2988 in the case of an équation x isn't a variable it's a constant
By differentiating u'll get 0=0
answer is about -0.703467