Solve this equation for S

[phantasm]

I'm using math for its main purpose in life...video games....but I can't seem to figure out how to solve this equation for S so I can easily graph some values. I need your help.


100 * log(( 3*S + 2*I + C) / 30 ) / log(2*M) = 100

[Khariz]

Yollia, you can handle this one.

[Sean of the Thread]

Bobmuthol is good at that shit hit him up for a solution.

[BriarFox]

I'm using math for its main purpose in life...video games....but I can't seem to figure out how to solve this equation for S so I can easily graph some values. I need your help.


100 * log(( 3*S + 2*I + C) / 30 ) / log(2*M) = 100

Huh. What's it for? It's odd that the whole log...M sequence is just going to equal 1 no matter what.

Edit: Oh, unless that 100 only applies to the top half of the equation?

[Xaerve]

assuming I is infinity:
S = -(2)(NTIF)/3 +/- 2M/3

assuming I is a variable:
S = [-(2x)/3] - [c/3] +/- 20M

[Trouble]

OK I haven't done math in 15 years but this is what I came up with:

s = 5/m - 2i/3 -c/3

[phantasm]

Its a formula that yields a %chance to cast.

S = Spell Rank
I = Int stat
C = Spell Craft skill ranks
M = Manacost of spell

I want to solve for different training plans that will still give 100% chance to cast so I set the equation equal to 100.


[100 * log(( 3*S + 2*I + C) / 30 ) / log(2*M) ] = 100

I'll manually switch the M and C variables, but I need to have it in Y=X form so I can easilly graph them in WZGrapher.

[ElvenFury]

My math is also rusty. So I ended up with S = (60m - C - 2i)/3

I'm assuming that I'm wrong though.

[phantasm]

Okay, I'll test these out and see if any of them give the same results as the original. I'm also curious what identities you use to get the variables out of the logarithm.

[Xaerve]

Mine answers be right.

[phantasm]

assuming I is infinity:
S = -(2)(NTIF)/3 +/- 2M/3

assuming I is a variable:
S = [-(2x)/3] - [c/3] +/- 20M


Woot, Xaerves's equation is giving proper results. Thanks. Assuming I is a variable and you mistyped it as x.

[ElvenFury]

Mine answers be right.

Where did you get X from?

[Xaerve]

Replaced the variable I with the variable X; I, when dealing with Logs, usually refers to Infinity.

Also, please post copious logs of your results!

[phantasm]

Are you going to post how you solved it. I'd really like to know.

[ElvenFury]

Replaced the variable I with the variable X; I, when dealing with Logs, usually refers to Infinity.

Also, please post copious logs of your results!

Ok, gotcha. Then our answers are pretty much the same thing, you just simplified M a bit, right?

(I was also confused by i at first)

[Xaerve]

I could scan my paper I guess, I'm not typing all that out!

EF: Yes, simplified M.

[Sean of the Thread]

Will you take my online math courses next semester?

I'm sure we can work something out.

[ElvenFury]

100 * log(( 3*S + 2*I + C) / 30 ) / log(2*M) = 100

Divide both sides by 100 to get:

log[(3s + 2i + C)/30] / log2m = 1

Multiply both sides by Log2m:

log[(3s + 2i + C)/30] = log2m

Get rid of the logs:

(3s + 2i + C)/30 = 2m

*30:

3s + 2i + C = 60m

Subtract the remaining junk and divide by 3 to get:

s = (60m -2i - C)/3

"Simplify" to:

s = 20m - (2i/3) - (c/3)

That's how I did it, anyway.

[Xaerve]

Thanks for saving me the time :)

Nice work.

[Sean of the Thread]

You fuckers make me sick to my stomach.

[ElvenFury]

It's been so long that I had to look up logarithms because I forgot what the hell they did. Luckily they ended up canceling each other out anyway. LOL

Xaerve: Thanks, and you're welcome.

Phantasm: You're welcome.

Sean: Sorry.

:-)

[phantasm]

Thats awesome thank you.

[Euler]

nerds.

[Bobmuhthol]

lol, thanks for believing in me, Sean. I wish I saw this thread sooner.