how do you cancel a 3/2 power?
can u give an example...not quite sure what u meant
You don't just cancel a 3/2 power. What exactly are you trying to do?
like how you would use a square root to cancel something squared
\[v ^{3/2} for instance\]
Well just to make sure you know what a 3/2 power is, 2^(3/2) = sq root (2^3)
to me its the same as this:\[\sqrt{v^{3}} \]
yes that's right
thats what i was looking for/ i dont think its going to do what i had intentions for though
You are probably looking for something like this right? 2^(3/2) = sq root (8) squaring both sides 8 = 8
[x^(3/2)]^2 = x^3
\[v _{y}^{3/2}=mg/b(1-e ^{(-b/m)t})\]
i was trying to get vy by itself
wow - that's a bit difficult. You could square both sides and you'd have vy^3 = [mg/b(1−e(−b/m)t)]^2 which might make an even bigger mess.
yeah thats what i was thinking. ill probably leave it the way it is. i had to find a equation of the terminal speed under a drag force that was proprtional to v\[_{y ^{3/2}}\]
Put both sides to the power of the reciprocal of the power. \[\huge v^{\frac{ 3 }{ 2 }} = x\] put both sides to the power 2/3 \[\huge (v^{\frac{ 3 }{ 2 }})^{\frac{ 2 }{ 3 }} = x ^{\frac{ 2 }{ 3 }}\] which gives \[\huge v = x ^{\frac{ 2 }{ 3 }}\]
That's how you get rid of any power - put both sides to the reciprocal of the power.
thank you sir, thats exactly what i was looking for
4(3/2) = 8 So, 4 = 8^(2/3) and it does Good work agent smith
:) well, that's the same thing you're doing when you take the square root of both sides of an equation - putting both sides to the power of 1/2, to get rid of a power of 2.
thats a good explanation
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