We fold the paper n times, so its thickness becomes
Since 0.1mm=10−7km,
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.