whats the formula for relating exponents? if u=y^-3 and y=u^3,

if u = y^-3 and y=u^3 then u*y^4=u^3
so u^2 = y^4.

so u^-2 = y^-4

[math]y^{-4} = (y^{-3 }) ^ { \frac{4}{3} }= u^\frac{4}{3} [/math]

>>8277079
This doesn't make any sense, because quite obviously [math] u^2 = y^{-6} \neq y^4 [/math]

>>8277079
is it like u*y^-3=y*u^3 and then y^-4 = u^2?

>>8277091
but the answer is u^2 i think

>>8277068
Dumbasses it's obviously [math]u^4[/math].

>>8277096
No.
>y^-4 = u^2
Does not follow from
>u*y^-3=y*u^3

>>8277098
But it's not. Try plugging in some numbers as an example and you will easily disprove that. It's not u^4 either.

The best i can think of is that taking away a y also takes away a u

u^2=y^-4
u=y^-3
1=y^-2
u^-1 = y^-1
u^-2 = 1
u^-3 = y

i think my problem was i forgot to account for y^0

but i think this is cheating because i already knew that if given u=y^-3 then y=u^3.

i wonder if there is a formula for this or something.

>>8277108
i mean taking away a y adds a u

I think it's either u^(4/3) or u^-12

Given if u=y^-3
Multiply both sides by y^-1
u * y^-1 = y^-4

Since y = u^3, y^-1 = u^-3 so we substitute
u * u^-3 = y^-4

u^-2 = y^-4

Did I make a mistake?

>>8277108
oh this doesnt work eitiher :(

>>8277113
oh that makes a lot of sense

u^-2 could have been the answer instead of u^2

thanks :D

>>8277095
it'll all make sense when both u and y turn out to be equal to 1

>>8277108
>i wonder if there is a formula for this or something.
Sure there is.
When [math] u = y^a [/math], [math] y^b = u^{ \frac{b}{a} } [/math].

>>8277068
I just realized the statement of the problem is wrong. You can't have
>u=y^-3 and y=u^3
in general.

u = y^-3 gives y=u^-1/3.

I think OP must've been working a problem that started somewhere further back, and ended up conflating negative exponentiation with fractional exponentiation.

To be clear, you are implying that one over the cube is equal to the cube root, which is obviously not true:

Say y=2, then u = 1/8. But (1/8)^3 certainly does not equal 2. (!!)

>>8277113
The mistake is here:
>Since y = u^3, y^-1 = u^-3 so we substitute
You can't have that when you also have
>Given if u=y^-3

The OP post just contains an incorrect specification.

>>8277132

u = y^-3 gives y=u^-1/3.

this was definetly the issue, thanks

it was a DE berenoulli problem, i always have trouble switching between y and u :(