Math/Science word Problem?

Question

Assume the thickness of a cell to be 0.1mm. (Although this is very thin, it is still much thicker than the average cell.) At any rate, assume that each thime the cell doubles, all of the new cells are stacked on top of one another. How many doublings would it take to produce a stack of cells that would reach to the moon (about 386,232 km away)?

Please Show Work.

Answer 1

OK...the moon is 386,232 km away, which is 386,232,000 m, which is 386,232,000,000 mm.

The cell is 0.1mm thick, so in the 386,232,000,000 mm you would fit 386,232,000,000 / 0.1 = 3,862,320,000,000 cells

With each doubling, the number and therefore height of cells doubles.
So with 1 doubling, you get 2^1 = 2 cells
With 2 doublings you get 2^2 = 4 cells
So you need to answer the question,
2^x = 3,862,320,000,000

What is x?

The simplest method is to take logs
log (2^x) = log 3,862,320,000,000
x log 2 = log 3,862,320,000,000
x = log 3,862,320,000,000 / log 2
= 41.81

At 41 doublings the stack will not reach the moon.
So you need 42 doublings for the stack to reach the moon.

Answer 2

I believe you are asking how many terms there are in the seies 1+ 2 +2^2+ 2^3 + 2^4 + 2^n + 2(n+1) + ........^, such that the sum multiplied by .1 mm = 386,232 km.The sum of the series for the first n terms is 1/(1-2) -2^n/(1-2) = -.5 + 2^n
So 0.1(-,5+2^n) = 386,232 times 10^5
-.5+2^n = 386,232 times 10^6
2^n = 3.86232 times 10^11 - 0.5
ln 2^n = ln (3.86232 times 10^11 - 0.5)
n ln2 = ln (3.86232 times 10^11 - 0.5)
n = [ln (3.86232 times 10^11 - 0.5)]/ln2 doublings

Answer 3

f(0) = 0.1 x 2^0
f(1) = 0.1 x 2^1
...
f(n) = 0.1 x 2^n

386,232 km = 386,232,000 m = 386,232,000,000 mm

So plug in 386,232,000,000 for f(n) and then solve for n...

386,232,000,000 = 0.1 x 2^n
Multiply both sides by 10:
2^n = 3,862,320,000,000

Now you would take the log base 2 of both sides:
log[2] 2^n = log[2] (3,862,320,000,000)
n = log[2] (3,862,320,000,000)

To convert bases to something you can do on your calculator, use the change of base formula:
log[a](b) = log(b) / log(a)

n = log(3,862,320,000,000) / log(2)
n ≈ 41.8 times

So after 42 folds, it would be past the moon.

Answer 4

386232km = 386232000m
0.1mm = 0.0001m

number of cells = 386232000/0.0001
= 3.863 x10E12

This number of cells would take 42 doubles as 2^42 = 4.398x 10e12

Answer 5

42 doublings.

.1mm x 1,000,000 = 1km
1,000,000 x 386232 = 386232000000 number of cells needed.

0.1
0.2
0.4
0.8
1.6
3.2
6.4
12.8
25.6
51.2
102.4
204.8
409.6
819.2
1638.4
3276.8
6553.6
13107.2
26214.4
52428.8
104857.6
209715.2
419430.4
838860.8
1677721.6
3355443.2
6710886.4
13421772.8
26843545.6
53687091.2
107374182.4
214748364.8
429496729.6
858993459.2
1717986918.4
3435973836.8
6871947673.6
13743895347.2
27487790694.4
54975581388.8
109951162777.6
219902325555.2
439804651110.4