I'm retarded... basic math help needed

[Jumpem]

How do I write p = q^-a in terms of q?

I need q = p +/- something.

 

[PowerMac4Ever]

you need to use logs and stuff

 

[Jumpem]

Originally posted by: PowerMac4Ever
you need to use logs and stuff

Do I? I thought there was a rule to pull the -a down.

 

[Mo0o]

q= p^(-1/a)

 

[Jumpem]

Originally posted by: Mo0o
q= p^(-1/a)

Thanks🙂

 

[Tweak155]

to show a bit more work, q^-a = 1/q^a

now, 1= p(q^a), q^a=1/p, then you get 1/p = p^-1

now to find the a'th root of p, add it to the denomintor of the power giving: p^(-1/a)

 

[Epoman]

Originally posted by: Tweak155
to show a bit more work, q^-a = 1/q^a

now, 1= p(q^a), q^a=1/p, then you get 1/p = p^-1

now to find the a'th root of p, add it to the denomintor of the power giving: p^(-1/a)

uhhhhh. yeah that's right.....😱

 

[Tweak155]

Originally posted by: Epoman

Originally posted by: Tweak155
to show a bit more work, q^-a = 1/q^a

now, 1= p(q^a), q^a=1/p, then you get 1/p = p^-1

now to find the a'th root of p, add it to the denomintor of the power giving: p^(-1/a)

uhhhhh. yeah that's right.....😱

lol, step by step:

p=q^-a
p=1/(q^a)
(q^a)p=1
(q^a)=1/p
(q^a)=p^-1
(q^a)^(1/a)=(p^-1)^(1/a)
q=p^(-1/a)

 

[Jumpem]

Thanks again guys.

As embarassing as it is to say I've done this all a million times. However, it's been years since I've used any of my college math courses so it's all been forgotten.🙁

 

[Epoman]

Originally posted by: Tweak155

Originally posted by: Epoman

Originally posted by: Tweak155
to show a bit more work, q^-a = 1/q^a

now, 1= p(q^a), q^a=1/p, then you get 1/p = p^-1

now to find the a'th root of p, add it to the denomintor of the power giving: p^(-1/a)

uhhhhh. yeah that's right.....😱

lol, step by step:

p=q^-a
p=1/(q^a)
(q^a)p=1
(q^a)=1/p
(q^a)=p^-1
(q^a)^(1/a)=(p^-1)^(1/a)
q=p^(-1/a)

uhhhhh. yeah that's right.....😱