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Edit: Thanks to Verdruss, Epiousios, and geeky me for solving this for me and teaching me about logarithms to boot! You've helped restore my sanity, it's been driving me crazy for over a week now.

I have variables A, B, and C. I'm trying to build a formula for any given value of A and B, such that it gives C in the patterns shown below. I'll start off with some examples.

If A=1, and B=50, then C=50
A=10, B=50, then C=100
A=100, B=50, then C=200
A=1000, B=50, then C=400
A=10000, B=50, then C=800

And so on. Or,

A=1, B=1, C=1
A=10, B=1, C=2
A=100, B=1, C=4
A=1000, B=1, C=8
A=10000, B=1, C=16

Ad infinitum.

A=1000, B=999, then C=7992

But what if...

A=6923, B=30, C=??? 

It would be somewhere between 240 and 480, but what exactly?

Every time A increases an order of magnitude, C doubles, with B being the starting value of C given A=1. I can solve this equation as long as I plug in 1, 10, 100, 1000, 10000, and so on into it, but not any other number for A. Now, I had an engineering friend help me a bit, but his solution gives me stratified, tiered equations bracketed by the magnitudes, and not say, a single sloped line per any constant B.

The equation I'm looking for would be something along the lines of ab=c, but I can't for the life of me figure out how to express the relationship in the above charts into mathematical terms.

And on top of it all, I'm pretty much a math amateur. I've done about 20 hours of Khan Academy, and tons of browsing through the different topics to find the one topic that will help me solve this specific equation I'm looking for. But most of the education I'm finding is related to solving equations, not building equations to further solve an equation.

I fear I'm in straight up Calculus territory here, but I don't know. Or maybe this equation is impossible to solve, despite how elegant it seems in simple terms.

The most complicated insight I could give into would be something along the lines, like, each real number increment of A is worth less/(or more?) then the number before it. IE, If B= 10, then C=10 if A=1, C=20 if A=10, but C= ~16 (or is ~14?) if A=5.

Or maybe there is a bigger equation here beyond just the scope of the variables I'm working with?

I feel like there is something in the appreciation(or depreciation) of the value each point of A is worth that is key to solving this.

Is this a trigonometric equation? Quadratic? Simple exponential? Multivariable calculus? Differentials? Are there math terms I've simply never seen before that could build this equation? Is there a type of mathematics that covers this kind of equation so I can research it?

...Should I give up? Or will I find, for sure!, the answer after I pour one thousand hours into learning advanced mathematics? Or is the equation truly impossible to solve/build?

Or that the simple A B C format doesn't hold on it's own until I figure out the bigger formula it is a part of? (Like maybe, A=1, B=X, C=X is impossible, but part of another equation the relationship between A and C can exist? (Where C doubles for every magnitude increase of A(And I will also concede that it might be how I've worded it, I'm pretty certain the doubling doesn't hold between, say, A=9 compared to A=90)))?

Sorry this post is kind of a thought salad. I'm just not certain the best way to explain what I want and my thoughts on various aspects of it.

Thanks for any help or guidance! I appreciate it.

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    $\begingroup$ Didn't read the whole post but what about $c=b2^{\log_{10}A}$? $\endgroup$ – Verdruss Oct 13 '17 at 10:13
  • $\begingroup$ in what kinda context you need this? I mean for the first part $A\cdot B=C$ if you consider $A$ written in binary. The second is also similar $A=C$ if you write $A$ in binary and $C$ in decimal. $\endgroup$ – Vinyl_cape_jawa Oct 13 '17 at 10:13
  • $\begingroup$ @Vinyl The context is for a play by post RPG. I like making my own mechanics, and daydream about formulas I could use. So it's for nothing important whatsoever, really. Only for a hobby. $\endgroup$ – brainfriiz Oct 13 '17 at 10:22
  • $\begingroup$ @Verdruss I'll have to learn logarithms tomorrow and test it! $\endgroup$ – brainfriiz Oct 13 '17 at 10:23
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The relationship is

$$ C = 2^{\log_{10}A}B. $$


Here's how to get there:

Note that $C$ always gets multiplied by $2$ upon an increase in $A$. This suggests a relationship of the form

$$ C=2^{f(A)}\times\text{something}, $$ where $f$ is some function that depends on $A$.

Now, each time $C$ is doubled, $A$ has been multiplied by $10$. So $f(A)$ must increase by $1$ whenever $A$ is multiplied by $10$. This is true for $$ f(A) = \log_{10}A. $$

So our guess is $$ C=2^{\log_{10}A}\times\text{something}. $$

Since for $A=1$, $\log_{10}A$ is zero so that $2^{\log_{10}A}=1$ we can conjecture

$$ C=2^{\log_{10}A} B, $$

which appears to work.


Since you said that you don't know about logarithms yet:

$$ \log_x y = z $$ means that $$ x^z = y, $$ so that $z$ is the power of $x$ which is equal to $y$. For example, $\log_2 16 = 4$ because $2^4=16$ and $\log_{10}100 = 2$ because $10^2=100$

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  • $\begingroup$ I don't know any logarithms, but now I know what to learn next! If this is truly correct, then.. wow. Thank you so much! It's super late here so I'll have to test it tomorrow, just spent the last 4 hours trying to figure this out with a friend in a video game and it's way passed my bedtime. $\endgroup$ – brainfriiz Oct 13 '17 at 10:21
  • $\begingroup$ "note that B always gets multiplied by 2" - shouldn't this be C? $\endgroup$ – Jam Oct 13 '17 at 10:30
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    $\begingroup$ @Jam Thank you, corrected this. $\endgroup$ – Epiousios Oct 13 '17 at 10:32
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Its simply by observing,
$$C=B * 2^{log_{10} A}$$

How did I think so??
There was an exponential (type of) increase on both the sides. And therefore, I thought it had something to do with logarithms and $2^x$ and in the first try I got the answer.

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  • $\begingroup$ Thanks for the reply! I'll have to learn logarithms and test it for myself, starting tomorrow. $\endgroup$ – brainfriiz Oct 13 '17 at 10:26
  • $\begingroup$ Once you learn that you will come to know how easy your question was.. BTW you got something to learn about. By the time you learn them -logarithms are just reverse of exponents. $\endgroup$ – geeky me Oct 13 '17 at 10:29

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