I went through a lot of the algebra, recalling some old log rules. But, I keep hitting dead ends. Also, I think what s confusing people the most is that the word "formula" should be "function". He's asking for a function that:

 

f(x) = xp remaining

 

 

 

Where function f is how much xp remains for given xp (x).

 

 

 

Here's what I have so far:

 

 

 

ln(4y-x) - ln(300) = (x/7)(ln(2))

 

 

 

(I tested it by plugging in numbers from the original function, and it's correct)

 

 

 

All I need to do is separate the x and y in the first logarithm....Any suggestions?

y = (x+300*2^(x/7))/4

 

 

 

To find x explicitly, you need the Lambert W function ( see http://en.wikipedia.org/wiki/Lambert%27s_W_function )

 

 

 

when y = x exp(x)

 

then x = W(y)

 

 

 

Your problem is an instance of the general equation

 

p^(a*x+B) = c*x+d

 

 

 

Specificaly:

 

y = (x+300*2^(x/7))/4

 

y = x/4+75*2^(x/7)

 

y/75 - x/300 = 2^(x/7)

 

 

 

thus c=-1/300 , d=y/75 , b=0 , a=1/7 , p=2

 

The wiki page I linked to above describes how to solve this kind of equation, the solution to your problem is

 

x = (-7*W(300/7*ln(2)*2^(4/7*y))+4*y*ln(2))/ln(2)

 

 

 

 

 

Unfortunately, Microsoft Excel does not contain an implementation of the Lambert W function. You can, however, approximate it to arbitrary accuracy using the Taylor series, also described in the wikipedia article I linked to.

y = (x+300*2^(x/7))/4

 

 

 

To find x explicitly, you need the Lambert W function ( see http://en.wikipedia.org/wiki/Lambert%27s_W_function )

 

 

 

when y = x exp(x)

 

then x = W(y)

 

 

 

Your problem is an instance of the general equation

 

p^(a*x+B) = c*x+d

 

 

 

Specificaly:

 

y = (x+300*2^(x/7))/4

 

y = x/4+75*2^(x/7)

 

y/75 - x/300 = 2^(x/7)

 

 

 

thus c=-1/300 , d=y/75 , b=0 , a=1/7 , p=2

 

The wiki page I linked to above describes how to solve this kind of equation, the solution to your problem is

 

x = (-7*W(300/7*ln(2)*2^(4/7*y))+4*y*ln(2))/ln(2)

 

 

 

 

 

Unfortunately, Microsoft Excel does not contain an implementation of the Lambert W function. You can, however, approximate it to arbitrary accuracy using the Taylor series, also described in the wikipedia article I linked to.

 

...what?

...what?

 

This is the solution:

 

x = (-7*W(300/7*ln(2)*2^(4/7*y))+4*y*ln(2))/ln(2)

 

where W is the Lambert W function.

 

that's it.

I know what to the is, didn't know you could write it with a ^ though

Don't worry guys, I solved it through a much easier method of just using VLOOKUP and copying an XP Table.

 

 

 

 

 

All I need now is an Excel function that can retrieve info directly from sites. If that even exists...

oh god I've forgotten math, Darn you runescape!

Bump for an Excel command that can extract data directly from sites.

 

 

 

Can anyone help?

Luls UW, you're such a nerd. :D

y = (x+300*2^(x/7))/4

 

4y = (x+300*2^(x/7))

 

4y-x = 300*2^(x/7)

 

(4y-x)/300 = 2^(x/7)

 

log2 of [(4y-x)/300] = x/7

 

x = 7{log2 of [(4y-x)/300]}

 

*the {} must be there. Do not simplify the 7 into the log.

 

 

 

And no garuntee I did that right. I think it works but its been a summer and a bit since I did logarithims.

 

 

 

EDIT:

 

Don't worry guys, I solved it through a much easier method of just using VLOOKUP and copying an XP Table.

 

 

 

 

 

All I need now is an Excel function that can retrieve info directly from sites. If that even exists...

 

There are javascript or html funtions that can retrive that sort of data, but the best you could do would be a script that takes that data and gives an output in the same way as the tip.it calcs. Excel cannot activly look up info from the internet.

 

x = 7{log2 of [(4y-x)/300]}

 

 

You still have an x on the other side of the equation. This is basically what I got.

 

ln(4y-x) - ln(300) = (x/7)(ln(2))

 

If its excel you want, use this forumla:

 

 

 

Cell A1: 1

 

Cell B1 formula: =ROUND.DOWN(A1+300*POWER(2;A1/7);0)/4

 

Cell B1: 2

 

Cell B2 formula: =B1+ROUND.DOWN(A2+300*POWER(2;A2/7);0)/4

 

 

 

Fill the rest of the first column with numbers 3, 4, ...

 

Copy B2 field formula down.

 

 

 

It's from RSDemon, but I have a similar one in Excel that works fine. Can't be bothered to go to it though :wall: