lol i forgot how to do math

[chasin225]

Im in a realy dumb school now, i realy forgot how to do maths haha. but ok i need a formula wich increases a number whens its bigger

example:

when i got number 80 i need it to be like 100
when i got number 3000 i need it to change to like 20000.

what formula is good for someting like this.

 

[Tozza]

Can you explain a little more of what you're needing.

 

[chasin225]

mmm ok i can remember a few years ago when i had math that there where formula's that higher increases a number when the number is bigger.

example:
If theres a formula that goes A * 3 = B
when A = 5 it goes 5 * 3 = 15
when A = 10 it goes 10 * 3 = 30

Well if i do 10 / 5 its 2
And 30 / 15 its also 2 and i want it to be different.

The formula i need should be different.
Something like

when A = 5 it goes 5 * 3 = 15
when A = 10 it goes 10 * 3 = 80 (this answer would fit better for me, but it doesent work with a formula that goes with only *3)

Hard to explain. hope you understand what i mean Its just basic math. Maybe someting with cos, sin, tan, log or whatever

 

[Sevion]

10 * 3 doesn't = 80 >_>

If I'm understanding you right, you're looking for either an exponential or power model.

AX^B or AB^X format.

 

[Super-Sheep]

Ye I think he means exponential if thats the word.

But now he wants it to be used in Warcraft III. But for what purpose would you need that? Because maybe you could use an alternative way to get around your problem.

 

[ap0calypse]

when i got number 80 i need it to be like 100
when i got number 3000 i need it to change to like 20000.

what formula is good for someting like this.

For that exact formula (with those numbers):

(A * 6.815) - 445.2 = B

If A = 80, then B = 100.
If A = 3000, then B = 20000.

(Rounded up, that is).

when A = 5 it goes 5 * 3 = 15
when A = 10 it goes 10 * 3 = 80 (this answer would fit better for me, but it doesent work with a formula that goes with only *3)

(A * 13) - 50 = B

A = 5: 5*13-50 = 15
A = 10: 10*13-50 = 80


There are a lot of formula's, but why do you need that "3"?

I mean, you can do: (A*13) - (16.67*3) = B
Then you've got your 3, but it's useless (as 16.67 * 3 = 50).