p. 482

83.

Find all solutions of the equation (x^2) = (2^x).

Plot[{(x^2),  (2^x)},  {x, -2, 5}, 
      PlotStyle -> {RGBColor[0, 0, 1], RGBColor[0, 1, 0]}]

The graphs show three crossings, at 2 and 4 and a number near -0.8. Of course, (2^2) =  (2^2) =  4 and (4^4)  =  (2^4) =  16. We use NSolve[] to find a numberical approximation to the third number.

Needs["Miscellaneous`RealOnly`"];

solns  =  NSolve[(x^2) ==  (2^x), x]

(2^x) /.  solns[[1]]

x^2  /.  solns[[1]]

Another good numerical function for solving problems like this is FindRoot[]. It is a bit more difficult to use than NSolve[] since you must give it a starting guess (the second parameter).

FindRoot[(x^2) ==  (2^x), {x, -1}]

84.

Could (x^(ln (2)))possibly be the same as (2^(ln (x)))for x > 0 ?

Plot[{(x^Log)[2],  (2^Log)[x]},  {x,  .1, 100}]

The graphs appear to be identical. We can verify that they are the same function by using the definition (a^b) =  (e^(b ln (a))):

\(x\^\(Log[2]\ \) == \ E\^\(Log[2] Log[x]\); 

    2\^Log[x] == \ E\^\(Log[x] Log[2]\); \)