Dividing by 2, then 2, then how to do this easily?

[ISX]

Dividing by 2, then 2, then...how to do this easily?

Homework Statement

The radius of a copper atom is roughly 1.3 x 10^-10m. How many times can you divide evenly a piece of 10cm copper wire until it is reduced to two separate copper atoms?

Homework Equations

No clue

The Attempt at a Solution

1.3 x 10^-10 is the radius of one atom and it wants the width of 2 full atoms so (1.3 x 10^-10)(4) = 5.2 x 10^-10. But that is in meters so x100 makes it cm or 5.2 x 10^-8.

Now if you take 10cm and divide it by 2, 29 times, you get 5.2 x 10^-8 , which is the same answer as in the back of the book. So you take 10/2 = 5/2 = 2.5... but how do you do this easily? The book makes no mention of anything even close to dealing with it. I'm thinking its some logarithm thing or I have no idea.

 

[Ray Vickson]

Homework Statement

The radius of a copper atom is roughly 1.3 x 10^-10m. How many times can you divide evenly a piece of 10cm copper wire until it is reduced to two separate copper atoms?

Homework Equations

No clue

The Attempt at a Solution

1.3 x 10^-10 is the radius of one atom and it wants the width of 2 full atoms so (1.3 x 10^-10)(4) = 5.2 x 10^-10. But that is in meters so x100 makes it cm or 5.2 x 10^-8.

Now if you take 10cm and divide it by 2, 29 times, you get 5.2 x 10^-8 , which is the same answer as in the back of the book. So you take 10/2 = 5/2 = 2.5... but how do you do this easily? The book makes no mention of anything even close to dealing with it. I'm thinking its some logarithm thing or I have no idea.

What is stopping you from trying it using logarithms?

 

[ISX]

I don't know how to do it exactly.. Kinda like I've seen the power of logarithms, not sure how to use them. "As seen on tv, results of actually using it vary" lol

 

[symbolipoint]

Divide by 2, one time = /2¹
Divide by 2, two times = /2²
Divide by 2, three times = /2³
..
..
Enough for seeing the pattern?

You want to start with 10 centimeter and divide by 2 until the result is nearly or the same as 1.3 x 10^-10 meters.
To keep measurements in meters, 10 cm = 10.0x10^-2 meter.

Can you make the correct equation?

 

[ISX]

I get 10.0x10^-2/2^x=1.3x10^-10 But I'm not sure how to solve for that. This is all introductory stuff and I haven't dealt with many exponents.

 

[symbolipoint]

I get 10.0x10^-2/2^x=1.3x10^-10 But I'm not sure how to solve for that. This is all introductory stuff and I haven't dealt with many exponents.

Yes that is the correct equation. You want to find some whole number value for x. Intermediate Algebra teaches about logarithmic and exponential functions. If you have not studied them yet, then just try brute force.

 

[Mandelbroth]

I get 10.0x10^-2/2^x=1.3x10^-10 But I'm not sure how to solve for that. This is all introductory stuff and I haven't dealt with many exponents.

Brute force looks...painful. I hope I can help...

Very good. You have the right equation. Now, here comes the fun part. First, you'll want to isolate that 2^x. Then, it becomes a matter of "logarithms". The logarithm of a number is the exponent you would raise a number, called the base, to get the number you are, for lack of a better term, logarithming.

We use the notation $log_bx = y$, where b is the base, x is the number we are taking the logarithm of, and log denotes logarithm. This is essentially like saying $b^y = x$. In your case, we are using a logarithm of base 2. Using the typical notation, $log_22^x = x$. If you want x, take the log base 2 of both sides of your equation where 2^x is isolated, and you'll get an answer for x.

 

[ISX]

Well I don't know how to get my calculator to do anything other than base 10 but I found one on the net.

Anyhow I got 10.0x10^-2/1.3x10^-10 = 769,230,769.23

Then I did log base 2 of that number and got 29.518841. I assume they rounded to 29.

I'll have to figure out how to get my calculator to do it. The log button just does base 10.

 

[Mandelbroth]

Well I don't know how to get my calculator to do anything other than base 10 but I found one on the net.

Anyhow I got 10.0x10^-2/1.3x10^-10 = 769,230,769.23

Then I did log base 2 of that number and got 29.518841. I assume they rounded to 29.

I'll have to figure out how to get my calculator to do it. The log button just does base 10.

There is a "change of base formula". $log_bx = \frac{logx}{logb}$.

 

[ISX]

Oh that's fancy as all get out. Thanks for the help!

 

[Ray Vickson]

Well I don't know how to get my calculator to do anything other than base 10 but I found one on the net.

Anyhow I got 10.0x10^-2/1.3x10^-10 = 769,230,769.23

Then I did log base 2 of that number and got 29.518841. I assume they rounded to 29.

I'll have to figure out how to get my calculator to do it. The log button just does base 10.

You can do it without logarithms: just keep dividing by 2 until you get to where you want to be.

To speed this up, you can divide a couple of times by 2^10 = 1024, so in two such divisions you have divided by 2 a total of 20 times. If you divide by 1024 again you will see that the result is very slightly too small, (so 2^30 is a bit too big), but 2^29 is not big enough, so the answer is 2^30. (The actual power 29.518841 is not an integer; rounding down to 29 does not get you a large enough divisor, so you need to round up to 30.)