O/T Calling all MATHS geniuses

[Codner's Crackpipe]

If there's one thing I know about maths people it's that nothing excites them more than proving their rain man tendencies to a generally disdainful audience. So please, if somebody can help me out by expressing the following equation in terms of P then please do. I'm simply too thick and lazy to do it myself.

Y = ((100/P)^(360/D))-1

As a bonus you might avert a workplace rampage in the process...though this is only really a bonus if you happen to work in my office, which if you're clever enough to work this out is unlikely.

Much obliged.

 

[yxee]

I get P = 100*(Y+1)^(-D/360)

 

[Codner's Crackpipe]

Correct answer (I already had the value for P, just no idea how it was derived).

Thank you very much, take the rest of the day off.

 

[mejonaNO12 aka riskit]

I get P = 100*(Y+1)^(-D/360)

Yeah....errr...me too.

 

[Notters]

Y = ((100/P)^(360/D))-1
Y+1 = 100/p * 360/D
(360 (Y+1))/D = 100/P
P = 100D/(360 (Y+1))

It's a while since I've done any maths though!

 

[ForestRowSeagull]

Y = ((100/P)^(360/D))-1
Y+1 = 100/p * 360/D
(360 (Y+1))/D = 100/P
P = 100D/(360 (Y+1))

It's a while since I've done any maths though!

How did you get from line 1 to 2 ? Unless I'm getting the wrong end of the stick as I thought ^ meant 'to the power of' and you seem to have got rid of that I place of the *

 

[Notters]

How did you get from line 1 to 2 ? Unless I'm getting the wrong end of the stick as I thought ^ meant 'to the power of' and you seem to have got rid of that I place of the *

Ah, I didn't know that's what ^ meant. Carry on...

 

[ForestRowSeagull]

Am I being stupid here or do you not have to take a log of both sides to get rid of the unknown in the power ?

 

[Digweeds Trousers]

Not ashamed to say that none of this means anything to me at all. Way over my head.

 

[Seagull27]

Am I being stupid here or do you not have to take a log of both sides to get rid of the unknown in the power ?

That's what I thought...

 

[ForestRowSeagull]

I have a mate who got 100% in further maths so may just ask him :L My a-level clearly isn't good enough

 

[Barrel of Fun]

I got stuck/went wrong at...

Y = ((100/P)^(360/D))-1

2(360/D) = LOG(Y,100/P)

 

[Biscuit]

Anyone else remember when maths was numbers?

 

[34064 Fighter Command]

No, surely it's the ( 360/D ) th root of both sides of the equation that's required, or raising both sides of the equation to 1 / ( 360/D ) which is the same as ( D/360 ).

Then you invert both sides and it becomes the reciprocal power which is - ( D/360 ).

yxee is right and Notters has just got a grade F ........

 

[Seagull27]

I think it goes something like this:

Y = ((100/P)^(360/D))-1

Y+1 = ((100/P)^(360/D))

log(Y+1) = log((100/P)^(360/D))

log(Y+1) = (360/D)*log(100/P)

(D/360)log(Y+1) = log(100/P)

(D/360)log(Y+1) = log100 - logP

logP = log100 - (D/360)log(Y+1)

logP = log100 + (-D/360)log(Y+1)

logP = log100 + log(Y+1)^(-D/360)

logP = log(100*(Y+1)^(-D/360))

P = 100*(Y+1)^(-D/360)

Giving the result in the first post.

I have too much time on my hands...

 

[Badger]

I'm a bit rusty but I'll have a go:

Y+1 = (100/P) ^ (360/D)
(Y+1)^D = (100/P)^360
(Y+1)^(D/360) = 100/P
P*((Y+1)^(D/360)) = 100
P = 100/((Y+1)^(D/360))


....or what Seagull27 said

 

[34064 Fighter Command]

Anyone else remember when maths was numbers?

About Primary school I think ...........

 

[Manx Shearwater]

Perhaps we need a picture of Rachel Riley at this point?

It might help.

 

[sydney]

Ah, I didn't know that's what ^ meant. Carry on...

 

[ForestRowSeagull]

I get stuck or go majorly wrong around the point where is stop:
Y=(100/P)^(360/D) -1
Y+1=(100/P)^(360/D)
log Y+1= log(100/P)^(360/D)
log Y+1= (360/D)*log(100/p)
STUCK