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Perplex

Exponents & Logarithms (Lesson 3/4)

Logarithm algebra

1 / 14

Discussion

Jimmy has a piece of paper and he notices that when he folds it in half, its thickness doubles. He measures the initial thickness of the paper as ​0.1mm. He wonders how many times he would have to fold the paper to reach the moon, which he knows is ​384000km​ from Earth.

How many times does Jimmy need to fold the piece of paper to reach the moon?

Solution:

We fold the paper ​n​ times, so its thickness becomes

​
tn​=0.1mm×2n.
​

Since ​0.1mm=10−7km,

​
tn​=10−7×2n(km).
​

We need ​tn​≥384000km. By guessing:

​
n=10:t10​=10−7×210=10−7×1024≈1.02×10−4km
​

We see ​10​ folds is far too few. So we try:

​
n=100:t100​=10−7×2100=10−7×1.27×1030≈1.27×1023km
​

Clearly, we overadjusted, as ​100​ would be far too many folds. Now that we know Jimmy should make between ​10​ and ​100​ folds we make successive guesses:

​
n=20:t20​=10−7×220=10−7×1.05×106≈0.105km,\\n=30:t40​=10−7×230=10−7×1.07×1012≈107km\\n=40:t40​=10−7×240=10−7×1.10×1012≈1.10×105km.
​

At ​n=40, ​t40​≈110000​ km, still about a third of Jimmy's desired thickness of ​384000​ km. One more fold doubles it, which would not be enough, but two more folds will quadruple the thickness to about ​440000km:

​
t42​≈22×1.10×105=4.40×105km,
​

which exceeds ​384 000​ km. Hence Jimmy must fold the paper ​42​ times.


As you can see, we do not yet have a good way to solve problems in which we need solve exponentials with very large powers. In this lesson, we will learn a more systemic way to solve problems like Jimmy's.

Exponents & Logarithms (Lesson 3/4)

Logarithm algebra

1 / 14

Discussion

Jimmy has a piece of paper and he notices that when he folds it in half, its thickness doubles. He measures the initial thickness of the paper as ​0.1mm. He wonders how many times he would have to fold the paper to reach the moon, which he knows is ​384000km​ from Earth.

How many times does Jimmy need to fold the piece of paper to reach the moon?

Solution:

We fold the paper ​n​ times, so its thickness becomes

​
tn​=0.1mm×2n.
​

Since ​0.1mm=10−7km,

​
tn​=10−7×2n(km).
​

We need ​tn​≥384000km. By guessing:

​
n=10:t10​=10−7×210=10−7×1024≈1.02×10−4km
​

We see ​10​ folds is far too few. So we try:

​
n=100:t100​=10−7×2100=10−7×1.27×1030≈1.27×1023km
​

Clearly, we overadjusted, as ​100​ would be far too many folds. Now that we know Jimmy should make between ​10​ and ​100​ folds we make successive guesses:

​
n=20:t20​=10−7×220=10−7×1.05×106≈0.105km,\\n=30:t40​=10−7×230=10−7×1.07×1012≈107km\\n=40:t40​=10−7×240=10−7×1.10×1012≈1.10×105km.
​

At ​n=40, ​t40​≈110000​ km, still about a third of Jimmy's desired thickness of ​384000​ km. One more fold doubles it, which would not be enough, but two more folds will quadruple the thickness to about ​440000km:

​
t42​≈22×1.10×105=4.40×105km,
​

which exceeds ​384 000​ km. Hence Jimmy must fold the paper ​42​ times.


As you can see, we do not yet have a good way to solve problems in which we need solve exponentials with very large powers. In this lesson, we will learn a more systemic way to solve problems like Jimmy's.