c4countdown

Okay, I'm having a blank moment. It feels like there should be a very easy answer to this, but possibly not.

If I've got this set of data where a = 2^b, how do I determine the value of a if I'm given b? Is there an easy formula or do I have to use calculus or something like that? I can obviously do it in a loop with code (which is what I'm about to do, as it won't really ever have to do more than about a dozen iterations to get near enough to the answer to be usable, so it's not a huge overhead), but I wondered if there was a neater way and it bugs me that I should probably know how to do it...

That's odd, I've never really considered doing that without programming. There's probably an Egyptian technique somewhere, I'll have a look.

Kai Laddiman wrote:That's odd, I've never really considered doing that without programming. There's probably an Egyptian technique somewhere, I'll have a look.

Oh okay, I assumed I was being dim. I was just a bit annoyed looking at such a simple set of data (actually my real data wasn't quite so simple, but the same principle applied) that I couldn't work out a simple formula. Maybe there isn't one then.

Jon Corby wrote:Okay, I'm having a blank moment. It feels like there should be a very easy answer to this, but possibly not.

If I've got this set of data where a = 2^b, how do I determine the value of a if I'm given b? Is there an easy formula or do I have to use calculus or something like that? if there was a neater way and it bugs me that I should probably know how to do it...

Just trying to wind up these old cogs of mine and I think it's done like this:
Take logs of both sides:
log a = b*log 2
therefore
b=(log a)/log 2

Hope that helps, it's been a long long time!!!!!!!!!

Ian Fitzpatrick wrote:

Jon Corby wrote:Okay, I'm having a blank moment. It feels like there should be a very easy answer to this, but possibly not.

If I've got this set of data where a = 2^b, how do I determine the value of a if I'm given b? Is there an easy formula or do I have to use calculus or something like that? if there was a neater way and it bugs me that I should probably know how to do it...

Just trying to wind up these old cogs of mine and I think it's done like this:
Take logs of both sides:
log a = b*log 2
therefore
b=(log a)/log 2

Hope that helps, it's been a long long time!!!!!!!!!

Wow, you're a genius! That does help in that it shows me how it's done, makes me feel a bit better for not being able to work it out myself (despite being a little bit mathsy, I can't think I actually know how to use logarithms anywhere), but I'll actually leave my code as it is (with the loop) as it's much easier to follow and tinker with. Thanks though!

Jon Corby wrote:

Ian Fitzpatrick wrote:

Jon Corby wrote:Okay, I'm having a blank moment. It feels like there should be a very easy answer to this, but possibly not.

If I've got this set of data where a = 2^b, how do I determine the value of a if I'm given b? Is there an easy formula or do I have to use calculus or something like that? if there was a neater way and it bugs me that I should probably know how to do it...

Just trying to wind up these old cogs of mine and I think it's done like this:
Take logs of both sides:
log a = b*log 2
therefore
b=(log a)/log 2

Hope that helps, it's been a long long time!!!!!!!!!

Wow, you're a genius! That does help in that it shows me how it's done, makes me feel a bit better for not being able to work it out myself (despite being a little bit mathsy, I can't think I actually know how to use logarithms anywhere), but I'll actually leave my code as it is (with the loop) as it's much easier to follow and tinker with. Thanks though!

You said you wanted to calculate a given b, not b given a...

Michael Wallace wrote:You said you wanted to calculate a given b, not b given a...

I can't remember much about logarithms, but I think that means ln a = 2 * ln b => a = e^(2 * ln b). You might want to check that.

a = 2 ^ b

Michael Wallace wrote:You said you wanted to calculate a given b, not b given a...

Haha, oh yeah. I obviously meant the other way around.

Kai Laddiman wrote:

Michael Wallace wrote:You said you wanted to calculate a given b, not b given a...

I can't remember much about logarithms, but I think that means ln a = 2 * ln b => a = e^(2 * ln b). You might want to check that.

You could simplify that further;
a=e^(2*lnb)
=e^(lnb*2)
=(e^lnb)^2
=b^2

Not that that's any help to the OP.

Kai Laddiman wrote:

Michael Wallace wrote:You said you wanted to calculate a given b, not b given a...

I can't remember much about logarithms, but I think that means ln a = 2 * ln b => a = e^(2 * ln b). You might want to check that.

Definitely something wrong here, I'm not sure how you got the first expression. a=b^2 is obviously not consistent with a=2^b which you started with.

Ian Fitzpatrick wrote:

Jon Corby wrote:Okay, I'm having a blank moment. It feels like there should be a very easy answer to this, but possibly not.

If I've got this set of data where a = 2^b, how do I determine the value of a if I'm given b? Is there an easy formula or do I have to use calculus or something like that? if there was a neater way and it bugs me that I should probably know how to do it...

Just trying to wind up these old cogs of mine and I think it's done like this:
Take logs of both sides:
log a = b*log 2
therefore
b=(log a)/log 2

Hope that helps, it's been a long long time!!!!!!!!!

I'm also dredging up old maths memories, but can you do:

b=(log a)/log 2
b=log a (log to the base 2)

Very helpful of course.

Charlie Reams wrote:

Kai Laddiman wrote:

Michael Wallace wrote:You said you wanted to calculate a given b, not b given a...

I can't remember much about logarithms, but I think that means ln a = 2 * ln b => a = e^(2 * ln b). You might want to check that.

Definitely something wrong here, I'm not sure how you got the first expression. a=b^2 is obviously not consistent with a=2^b which you started with.

Oops, went wrong there, sorry. It's sort of irrelevant now that Corby's changed his initial question anyway.