p.504

21.

a) Show that ln (x) grows slower than (x^(1/n))for any positive integer n.

Let a  =  1/n. Then (lim_(x -> ∞))((x^a)/(ln (x)))  =  
        (lim_(x -> ∞)) ((ax^(a - 1))/(1/x)) = 
          (lim_(x -> ∞)) (ax^a)  =  ∞.

b) Find an x such that (x^(1/1, 000, 000))  >  ln (x).

Try x  =  (e^((10^a))). Then (x^(1/1, 000, 000)) =  (e^((10^(a - 6))))and ln (x)  =  (10^a). So we need to find a value of a such that (e^((10^(a - 6)))) >  (10^a). If a  =  6, (e^((10^(a - 6)))) =  e, which is not greater than (10^6).
If a  =  7, then (e^((10^(a - 6)))) =  (e^10)

N[(E^10)]

which is still not greater than (10^7). But when a  =  8, we get (e^((10^(a - 6)))) =  (e^100),

N[(E^100)]

which definitely is greater than (10^8). Therefore, x  =  (e^((10^8)))is a number such that (x^(1/1, 000, 000))  >  ln (x) .

N[(E^((10^8)))]

c) Find a value for x where the graphs of (x^(1/10))and ln (x) cross.

The text is a litle imprecise here. There are two points with x  >  1 where the graphs cross, one near 1 and the other a very large number. The text is asking you to find the latter. The smaller one is easy to find:

Plot[{(x^(1/10)), Log[x]},  {x, 1, 10}, 
      PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 0, 1]}]

They seem to cross near x  =  3:

FindRoot[(x^(1/10))  ==  Log[x],  {x, 3}]

To find the larger one, we can first examine the problem as in part a). Try x  = (e^((10^a))). Then (x^(1/10))  =  ln (x)becomes (e^((10^(a - 1))))  =  (10^a). We easily see that a  =  1 is too small, since (e^((10^0))) =  e  <  (10^1) . But a  =  2 is way too large: (e^((10^1))) =  22026.4657948067066`  >  (10^2). Therefore we should look for a solution between x  =  (e^10)and, say,  x  =  (e^50).  Some experimenting is necessary to be able to see where the graphs cross:

Plot[{(x^(1/10)),  Log[x]},  {x, (E^35), (E^36)}, 
      PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 0, 1]}]

Now we can use numerical methods to find a more exact answer:

FindRoot[(x^(1/10)) ==  Log[x],  {x, 3.5×(10^15)}]

Therefore, (x^(1/10))crosses above the graphs of ln (x) when x  ≥  3.43063 × (10^15).

It is interesting to carry this further and find where (x^(1/1, 000, 000))  =  ln (x). This is equivalent to ln (x)/10^6  =  ln (ln (x)). From part a) we know we should look between x  =  (e^((10^7)))and x  =  (e^((10^8))). These numbers are too big for Mathematica to work with numerically. Let's try putting y  =  ln (x) and plotting y/(10^6) and  ln (y) from (10^7)to (10^8) instead.

Plot[{y/(10^6), Log[y]},  {y, (10^7),  (10^8)}, 
      PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 0, 1]}]

FindRoot[ y/(10^6)  ==  Log[y],  {y, 2×10^7}]

The value of x  is thus

(E^y)  /.  %

a number with over 7 million decimal places. Even at 5000 digits per page, it would take 1,445 pages to print out.