Examples from Section 6.4
(a^x) and (Subscript[log, a]) x

p. 480

Derivatives

For any positive number a  =  (e^(ln (a))) so for any function f (x), (a^(f (x))) =  (e^(f (x) ln (a))). Then the chain rule gives (∂_x) (a^(f (x)))  =  
        (∂_x) (e^(f (x) ln (a))) =  
          (e^(x ln (a))) f' (x) ln (a)  =  
            (a^(f (x))) f' (x) ln (a).

Similarly, (Subscript[log, a]) (f (x))  =  ln (f (x))/ln (a) and the derivative is (∂_x) (Subscript[log, a]) (f (x))  =  
        (∂_x) ((ln (f (x)))/(ln (a)))  =  
          ((f' (x))/(f (x) ln (a))).

11.

y  =  (2^x);

D[y, x]

13.

y  =  (5^((Sqrt[s])));

D[y, s]

32.

y  =  Log[7,  ((Sin[t] Cos[t])/((E^t) (2^t)))];

D[y, t]

Simplify[%]

Logarithmic Differentiation

42.

y  =  (t^((Sqrt[t])))= (e^((Sqrt[t]) ln (t)));
(∂_t) y  =  
        (∂_t) (e^((Sqrt[t]) ln (t)  =  ))(e^((Sqrt[t]) ln (t) (∂_t) (Sqrt[t]) ln (t)))= (e^((Sqrt[t]) ln (t) 
          ((1/2) (t^(-1/2)) ln (t)  +  (Sqrt[t]) (1/t))
            )) = (t^((Sqrt[t]))) 
      ((1/2) (t^(-1/2)) ln (t)  +  (Sqrt[t]) (1/t))

D[(t^((Sqrt[t]))), t]

Simplify[%]

46. y  =  
        ln (((x))^(ln (x))) =  
          (e^(ln (x) ln (ln (x))));
(∂_x) y  =  
        (∂_x) (e^(ln (x) ln (ln (x)))) =  (e^(ln (x) ln (ln (x)))) (∂_x) ln 
          (x) ln (ln (x)) =  ln (((x))^(ln (x))) 
        ((1/x) ln (ln (x))  +  
            ln (x) (1/(ln (x))) (1/x))

D[Log[x]^Log[x],  x]

Simplify[%]

Integration

53.

(∫_0^(π/2))(7^(cos (t))) sin (t) 
          dt. Substitute u  =  cos (t), du  =  -sin (t) dt.

∫(7^Cos)[t] Sin[t] \[DifferentialD]t

(∫_0^(π/2))(7^Cos)[t] Sin[t] \[DifferentialD]t

67. (∫_0^9)(( 2  (Subscript[log, 10]) (x + 1))/(x + 1)) 
          dx. Substitute u  =  
        (Subscript[log, 10]) (x + 1)  =  
          ((ln (x))/(ln (10))), du  =  (1/(ln (10) (x + 1))) dx.

∫((2  Log[10, x + 1])/(x + 1)) \[DifferentialD]x

(∫_0^9)(( 2  Log[10, x + 1])/(x + 1)) 
        \[DifferentialD]x

Theory and Applications

79.

By what factor k do you have to multiply the intensity I of the sound from your audio amplifier to add 10 db to the sound level.

Solution. Let L be the origininal sound level. The equation on p.480 is:
L  =  10 (Subscript[log, 10]) (I × (10^12)).
We must solve for k in the following eqution:
L  +  10  =  10 (Subscript[log, 10]) (k I × (10^12)).
Exponentiating these equations to base 10 gives the following equations:
(10^(L/10))  =  I × (10^12)
(10^(L/10  +  1))  =  k I × (10^12)
Dividing the second by the first gives 10  =  k.

l  =  10 Log[10, i (10^12)];

Solve[l  +  10  ==  10 Log[10, k i (10^12)],  k]

Simplify[%]

Grapher Explorations

83. (Modified)

Find all solutions of the equation (x^4) = (4^x).

Plot[{(x^4),  (4^x)},  {x, -2, 4}, 
      PlotStyle -> {RGBColor[0, 0, 1], RGBColor[0, 1, 0]}]

Plot[{(x^4),  (4^x)},  {x, 3, 5}, PlotRange -> {200, 300}, 
      PlotStyle -> {RGBColor[0, 0, 1], RGBColor[0, 1, 0]}]

The graphs show three crossings, at 2 and 4 and a number near -0.7. Of course, (2^4) =  (4^2) =  16 and (4^4)  =  (4^4) =  256. We use NSolve[] to find a numberical approximation to the third number.

Needs["Miscellaneous`RealOnly`"];

solns  =  NSolve[(x^4) ==  (4^x), x]

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

x^4  /.  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^4) ==  (4^x), {x, -1}]