Examples from Section 6.3
The Exponential Function

p. 472

Solving Equations with Logarithmic or Exponential Terms

9.

Solve[Log[y  -  1]  -  Log[2]  == x  +  Log[x],  y] // Simplify

Step by step:

Log[y  -  1]  -  Log[2]  == x  +  Log[x]

Log[((y - 1)/2)]  == x  +  Log[x]

((y - 1)/2) == (E^(x + Log[x])) ==  x (E^x)

y =  1  +  2 x (E^x)

11.

a)

Needs["Miscellaneous`RealOnly`"]

Solve[(E^(2  k)) == 4]

Step by step :

(E^(2  k)) ==  4

2  k  ==  Log[4]  ==  Log[(2^2)]  ==  2 Log[2]

k  ==  Log[2]

b)

Solve[ 80 (E^k) ==  1,  k]

80 (E^k) ==  1

(E^k) ==  1/80

k  ==  Log[1/80]  ==  - Log[80]

c)

Solve[(E^(Log[0.8] k)) ==  0.8]

(E^(Log[0.8] k)) ==  0.8

Log[0.8] k  ==  Log[0.8]

k  ==  1

Derivatives

21.

(∂_x)((9 x^2  -  6  x  +  2) (E^(3  x)))

Simplify[%]

32.

(∂_t) Log[((Sqrt[t])/(1  + (Sqrt[t])))]

Simplify[%]

37.

We assume y is a function of x, y (x) :  ln (y (x)) ==  (e^(y (x))) sin (x).

(∂_x) Log[y[x]]  ==  
      (∂_x) ((E^y)[x] Sin[x])

Solve[ %,   y'[x]]

Integrals

46.

∫2  (e^((2  x - 1))) dx. Substitute u  =  2  x  - 1, du  =  2  dx:

∫2  (E^(2  x - 1)) \[DifferentialD]x  ==  
      ∫(E^u) \[DifferentialD]u  ==  (E^u) +  C  ==  
      (E^(2  x - 1)) +  C

∫2  (E^(2  x - 1)) \[DifferentialD]x

61.

∫((e^r)/(1 + (e^r))) dr. Substitute u  =  1  +  (e^r), du  =  (e^r) dr:

∫((E^r)/(1 + (E^r))) \[DifferentialD]r  ==  
      ∫(1/u) \[DifferentialD]u ==  Log[u]  +  C  ==  
      Log[1 + (E^r)]  +  C

∫((E^r)/(1 + (E^r))) \[DifferentialD]r

Theory and Applications

67.

Find the extreme values of f (x)  =  (e^x) -  2  x on [0, 1].

f[x_]  :=  E^x  -  2  x;

f'[x]

Solve[%  ==  0,  x]

Plot[f'[x],  {x, 0, 1}]

Since f' (x)  <  0 for x < ln (2) and f' (x)  >  0 for x  >  ln (2), the function f (x) decreases for x < ln (2)and increase for x  >  ln (2), making f (ln (2)) the absolute minumum of f (x) for all x. There is no maximum and f (x) grows without bound as x -> ∞ and x -> -∞.

f[Log[2]]

N[%]

N[Log[2]]

Plot[f[x],  {x, -1, 3}]

77.

The linearization of (e^x) at x  =  0.

f[x_]  :=  (E^x);

The linear approximation is also the equation of the tangent line:

y  =  f[0]  +  f'[0] (x  -  0)

\(Plot[{f[x], 1 + x}, {x, \(-1\), 1}, 
      PlotStyle -> {RGBColor[0, 0, 1], 
			RGBColor[0, 1, 0]}]\)

To estimate the error of approximation on [0, 0.2], Let's plot the difference on this interval:

Plot[f[x] - (1 + x), {x, 0, 0.2}]

In fact, the difference is always increasing and reachies its maximum at x  =  0.2, so the maximum error on this interval is

f[x] - (1 + x)  /.  x -> 0.2