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    • erplex

    Exponents & Logarithms (Lesson 2/4)

    Logarithm algebra

    1 / 29

    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 100folds 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≈107kmn=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.