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Very nice job.
Thank you so much! ❤
Nice problem and excellent explanation of using the W function to solve for x.
Thanks for liking! ❤
3×3^(3x)=3^2/x , 3^(3x-1)=2/x=k ,2/x=k , x=2/k then3^(3×2/k-1)=k , 3^[(6-k)/k then3=k^[k/(6-k)] , 3^[3/(6-3)]=k^[k/(6-k)] so. k=3x=2/k , x=2/3
Very nice, indeed.
Very nice! ❤
3×.(3^3)^×=2(3^2)W{3×.Ln3.e^ln3(3×)}=W{2ln3.e^ln3(2)}=>3×ln3=2ln3=>3×=2,×=2/3
(3^3×)=18/3×=6/×=>1/6=(×.3^-3×)=>-3/6=(-3×)Ln3.e^ln3(-3×)=>(-1/2)ln3==W{(-3×)Ln3.e^ln3(-3×.e^ln3(-3×)}
Very nice job.
Thank you so much! ❤
Nice problem and excellent explanation of using the W function to solve for x.
Thanks for liking! ❤
3×3^(3x)=3^2/x , 3^(3x-1)=2/x=k ,
2/x=k , x=2/k then
3^(3×2/k-1)=k , 3^[(6-k)/k then
3=k^[k/(6-k)] ,
3^[3/(6-3)]=k^[k/(6-k)] so. k=3
x=2/k , x=2/3
Very nice, indeed.
Very nice! ❤
3×.(3^3)^×=2(3^2)
W{3×.Ln3.e^ln3(3×)}
=W{2ln3.e^ln3(2)}
=>3×ln3=2ln3
=>3×=2,×=2/3
Very nice! ❤
(3^3×)=18/3×=6/×
=>1/6=(×.3^-3×)
=>-3/6=(-3×)Ln3.e^ln3(-3×)
=>(-1/2)ln3=
=W{(-3×)Ln3.e^ln3(-3×
.e^ln3(-3×)}
Very nice! ❤