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Thanks for the love of my channel
Ну, блин, зачем так сложно и так подробно?
x3⁻ˣ = 1/9(-xln3)e⁻ˣˡⁿ³ = -1/9ln3(-xln3)e⁻ˣˡⁿ³ = 1/9ln(1/3)(-xln3)e⁻ˣˡⁿ³ = (1/3²)(1/3)3ln(1/3)(-xln3)e⁻ˣˡⁿ³ = (1/3³)ln(1/3³)-xln3 = ln(1/3³)xln3 = ln3³ = 3ln3*x = 3*
-x*ln(3)=LambertW(-1/9*ln(3))x=LambertW(-1/9*ln(3))/-ln(3)x~0.127869...
just by looking at it, x=3 , test , 3^(3-2)= 3^1 , 3^1=3 , OK ,
You missed a solution, both are monotonically increasing functions so there are two solutions. This is why just solving by inspection is not sufficient
@@benyseus6325 Thank you for your comment, I entered the other root one line higher, it's true just to look at it, welcome
Lambert W , -> W(-ln3/9)= - 0.14047892 , - ln(9x)= - 0.14047892 , x=(e^0.14047892)/9 , x=~ 0.127869 , test , 3^(0.127869 - 2)=0.127869 , same , OK ,
х=3
It is obvious that x is 3!!!
Esta mal. Cuando reemplazo 3 por el log neperiano, quedaria con el -x un producto de potencias, no es un exponente
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Thanks for the love of my channel
Ну, блин, зачем так сложно и так подробно?
x3⁻ˣ = 1/9
(-xln3)e⁻ˣˡⁿ³ = -1/9ln3
(-xln3)e⁻ˣˡⁿ³ = 1/9ln(1/3)
(-xln3)e⁻ˣˡⁿ³ = (1/3²)(1/3)3ln(1/3)
(-xln3)e⁻ˣˡⁿ³ = (1/3³)ln(1/3³)
-xln3 = ln(1/3³)
xln3 = ln3³ = 3ln3
*x = 3*
-x*ln(3)=LambertW(-1/9*ln(3))
x=LambertW(-1/9*ln(3))/-ln(3)
x~0.127869...
just by looking at it, x=3 , test , 3^(3-2)= 3^1 , 3^1=3 , OK ,
You missed a solution, both are monotonically increasing functions so there are two solutions. This is why just solving by inspection is not sufficient
@@benyseus6325 Thank you for your comment, I entered the other root one line higher, it's true just to look at it, welcome
Lambert W , -> W(-ln3/9)= - 0.14047892 , - ln(9x)= - 0.14047892 , x=(e^0.14047892)/9 , x=~ 0.127869 ,
test , 3^(0.127869 - 2)=0.127869 , same , OK ,
х=3
It is obvious that x is 3!!!
Esta mal. Cuando reemplazo 3 por el log neperiano, quedaria con el -x un producto de potencias, no es un exponente