on desmos you can switch to logarithmic view in the x or y direction, doing so in the x direction, as we did with our substitution, lets us see all the intersections of the graph (in logarithmic view it is identical to x^1/3 and x/3 in linear view :P)
to clarify you can also see all of them in linear, but because two of them are very small and the other is relatively large it is better to use logarithmic view to visualize the scale
By law of logarithms: 1/3(lnx) = cbrt(lnx) lnx = t Cube both sides (1/27)(t^3) = t t^3 = 27t t^3 - 27t = 0 t(t^2 - 27) = 0 t = 0, lnx = 0, x = 1 t^2 = 27 t = ±√27, lnx = ±√27, x' = e^3√3, x'' = e^(-3√3) x = 1, x = e^3√3, x = e^(-3√3) All solutions
I like the sentiment that e^(sqrt(27)) is a very large value.It's about 180, which is nearly twice the boiling point of water in degrees C!
It's the difference between boiling water and freezing water in Fahrenheit.
@@mbmillermo We should all learn to respect numbers we can't count up to using our hands.
on desmos you can switch to logarithmic view in the x or y direction, doing so in the x direction, as we did with our substitution, lets us see all the intersections of the graph
(in logarithmic view it is identical to x^1/3 and x/3 in linear view :P)
to clarify you can also see all of them in linear, but because two of them are very small and the other is relatively large it is better to use logarithmic view to visualize the scale
You do substitution, and it's a t party.
This was too simple as compared to your other problems❤
By law of logarithms: 1/3(lnx) = cbrt(lnx)
lnx = t
Cube both sides
(1/27)(t^3) = t
t^3 = 27t
t^3 - 27t = 0
t(t^2 - 27) = 0
t = 0, lnx = 0, x = 1
t^2 = 27
t = ±√27, lnx = ±√27, x' = e^3√3, x'' = e^(-3√3)
x = 1, x = e^3√3, x = e^(-3√3)
All solutions
Muito bom!
1,e^3√3
X = 1, e^+- sqrt(27)
x = 1
x=0 or x=e^3√3
e^(-3sqrt(3))