doubling a number over and over

[rogerxd45]

ok so i want to know if there is an easier way to figure out how many times you would have to double a number before getting to certain number. (other then going 2x2x2x2x2x2x2

 

question two lets say we take the number 1080 is there a way to figure out how many times it had been doubled assuming the start point is 1

 

what if the starting number was 4 any way to figure that out besides the long way

 

and this is not homework, i was just folding a piece of metal on itself over and over (like Damascus steel but i used indium since its soft and doesn't require heat to get the layers to bond but it also doesnt add any strength i just wanted to see how many layers i could get. unfortunately my computer needed a reboot and i forgot to write down the number it was roughly 3.8^43 layers ) so each time i fold it the number of layers doubles, so it got me thinking about

 

interesting not about the number of layers in the indium, the indium was 0.8mm thick so based on those numbers i believe each layer would be thinner then an atom which i know it isnt possible to have layers thinner then the atoms in those layers. it like the thing one of my teachers said about if ia guy was waolking across the road but could only do it in halves (so at first he woud be in the middle of the road second he would be 3/4 of the way across...and so on. but he could never make it across the road. obviously not realistic the teacher made his point) but i did this like 6 months ago so i might have scred up the math on how many indium atoms thick 0.8mm would be. but am positive it was 3.8 to the 43rd

Edited April 23, 2012 by rogerxd45

[the tree]

ok so i want to know if there is an easier way to figure out how many times you would have to double a number before getting to certain number. (other then going 2x2x2x2x2x2x2

 

question two lets say we take the number 1080 is there a way to figure out how many times it had been doubled assuming the start point is 1

That would be what logarithms are for. If y=2^x then log₂(y)=x.

 

In the case of your example, log₂(1080)=10.077...

 

So we know that it's not 1 doubled a whole number of times. The best we can say at that point is 2¹⁰<1080<2¹¹.

 

A little further investigation will tell you that 1080=2¹⁰+56.

 

If your calculator cannot do base 2 logarithms then you'll have to use the natural logarithm ( ln or log_e ) and to do that you'll need to know that log_b(x)=ln(x)/ln(b).

 

what if the starting number was 4 any way to figure that out besides the long way

Well I think since 4=2² you might be able to work that out. Edited April 23, 2012 by the tree

[John Cuthber]

Very roughly, 10 doublings gives you a thousand (3 digits).

Twenty will give you a six digit number and so on.

If you want 10^43 then that's 43 digits that's about 143 doublings

My calculator tells me that 2^143 is 1.1X10^43

 

2^145 is 4.46 *10^43

So it's about 145 doublings.

An indium atom is about 300pm in diameter so 10^43 layers of them would be impossibly thick (3E34 metres, much bigger than the solar system).

 

A milimetre thick sheet would be "only" about six million atoms thick.

Two dozen foldings would mean the "layers" were just about an atom thick.

[phillip1882]

as the previous two posters said, logarithms is the way.

if you don't have a calculator, you can aproximate a logarithm using euler's method, which is quite lengthy.

continuously multiplying by the factor is faster. or a slide rule could be helpful.

[Cayman]

Just writing the solution a little differently in case it helps anyone...

 

I believe the number of times a smaller number needs to be doubled in order to equal a larger number = Log((larger number / smaller number), base2).

To prove it the other way, just multiply the smaller number by (2 raised to that solution).

 

Similarly, for example, if you wanted to know how many times you need to iteratively multiply a small number by 8 in order to equal a larger number, just use base 8 in the log funtion. The prove it the other way by multiplying the small number by (8 raised to that solution).

 

Hope that helps somebody.