13.6. Solved problems

1. Calculate the future value of an investment of $1000 which grows at a continuous rate of 15% for 30 years.

   A(30) = $1000 e0.10(30) = 1000 × 90.01713 = $90,017.13
   

2. A bacteria species has a growth rate of k = 0.03 [hr^-1] in a laboratory culture. For an initial population of 1000, calculate the population after one day, two days, and three days.

   Q(t)  = 1000 e0.03t
   
   Q(24) = 1000 e0.03(24) = 2054
   
   Q(48) = 1000 e0.03(48) = 4221
   
   Q(72) = 1000 e0.03(72) = 8671
   

3. The isotope U²³⁵ decays to Pb²⁰⁷ with a half live of 7.1 × 10⁸ years. T is the mean lifetime of the parent nuclei and is used in the exponential decay equation:

   N = No e-(t/T)
   

   Compute the value of T.

   e-7.1 x 108/T = 2-1
   
   7.1 × 108 
   ---------- = ln 2
       T 
   
        7.1 x 108
   T = ----------- = 1.024313479 x 109 [years]
          ln 2 
   

4. The mathematical expression that relates radioactive decay to geologic time is called the age equation:

       ln(1 + D/P)
   t = ----------- 
            T
   

   The parameters are defined as follows:

   t - is the computed age of the rock or mineral specimen
   D - is the number of atoms of a daughter product today
   P - is the number of atoms of a parent isotope today
   T - is the mean lifetime of the parent nuclei

   The isotope K⁴⁰ decays to Ar⁴⁰ with a half life of 1.3 × 10⁹ years. Measurements from a sample of a rock count 2000 daughter atoms (Ar⁴⁰) and 400 parent atoms (K⁴⁰). Compute the rock's age.

         ln 2
   T = ---------
       1.3 x 109
   
       ln(1 + 2000/400)   1.3 × 109 ln 6
   t = ---------------- = -------------- 
               T              ln 2
   
   t = 3.360 x 109 years
   

5. Astronomers define the absolute magnitude to be the apparent magnitude that a star would have if it were 10 parsecs (32.6 light years) from the Earth. Brightness obeys the inverse square law. The Sun has an apparent magnitude of -26.8. Compute its absolute magnitude using the stellar magnitude equation:

   m2 - m1 = 2.5(log b1 - log b2) = 2.5 log (b1/b2)
   

   One light year is approximately 9.46 × 10¹² kilometers. The Sun is about 1.496 × 10⁸ kilometers (1 Astronomical Unit) from Earth. Using the inverse square law, we get the brightness ratio:

   babs / b1 AU = ( 1.496 × 108 / 32.6 × 9.46 × 1012 )2 = 
   
              = (4.851 × 10-7)2 = 2.353 × 10-13 = 
   
   

   This can be used in the stellar magnitude equation as follows:

   m1 AU - mabs = 2.5 log (babs / b1 AU)
   
   mabs = m1 AU - 2.5 log (babs / b1 AU) = 
   
       = -26.8 - 2.5 log 2.353 × 10-13 = 
   
       = -26.8 + 31.6 = 4.8  is the absolute magnitude of the Sun
   

6. A horde of mice is accidentally introduced onto an island. The initial population is 2000. It is estimated that the long term sustainable population is 2,000,000 individuals. Calculate the population after one year, two years, five years, and ten years. Then calculate the time to reach 1,000,000 given a logistic growth function of:

   N(t)  = 2000000/(1 + 999e-0.2t)
   
   N(0)  = 2000000/(1 + 999) = 2000     at start
   
   N(1)  = 2000000/(1 + 999e-0.2(1))
         = 2000000/(1 + 817.91202)
         = 2442
   
   N(2)  = 2000000/(1 + 999e-0.2(2))
         = 2000000/(1 + 669.64973)
         = 2982
   
   N(5)  = 2000000/(1 + 999e-0.2(5))
         = 2000000/(1 + 367.51156)
         = 5427
   
   N(10) = 2000000/(1 + 999e-0.2(10))
         = 2000000/(1 + 135.199948)
         = 14684
   
   Compute time to reach 1,000,000 mice.
   
   1000000 = 2000000/(1 + 999e-0.2t)
         1 = 2/(1 + 999e-0.2t)
   
   1 + 999e-0.2t = 2
   
   1 = 999e-0.2t
   
   e0.2t = 999
    
   0.2t = ln 999
    
   t = 5 ln 999 = 34.5 years
   

7. Solve: e^2x - 4 = 0

   e2x - 4 = 0
   
   e2x = 4
   
   2x = ln 4 = 2 ln 2
   
   x = ln 2 = 0.693147
   

8. Solve: e^2x - 3e^x + 2 = 0

   e2x - 3ex + 2 = 0
   
   (ex - 1)(ex - 2) = 0
   
   ex = 1  or  ex = 2
   
   x = ln 1  or  x = ln 2
   
   x = 0  or  x = 0.693147
   

9. Solve: log_x 81 = 2

   x2 = 81
        __
   x = √81 = 9
   

10. Solve: log_x 4096 = 4

    x4 = 4096
    
    x = 40961/4 = 8
    

11. Solve: log_x 32 = 5

    x5 = 32
    
    x = 321/5
    
    x = 2
    

12. Solve: log₅ x = 4

    x = 54
    
    x = 625
    

13. Solve: log 100000 = x

    log 105 = 5 log 10 = 5 × 1 = 5
    

14. Solve: 12^x = 18

    log 12x = log 18
    
    x log 12 = log 18
    
         log 18    1.255272
    x = -------- = --------- = 1.163171
         log 12    1.079181
    

15. Rewrite as a sum of simple expressions: ln 2e^{5x + 2}

    ln 2e5x + 2 = ln 2 + 5x + 2 
    
             = 5x + 2.693147
    

16. Rewrite as a sum of simple expressions: ln (x² - 9)

    ln (x2 - 9) = ln (x + 3)(x - 3)
    
                = ln (x + 3) + ln (x - 3)
    

17. Solve: log x - log (x - 1) = 1

    log x - log (x - 1) = log 10
    
      x
    ----- = 10
    x - 1
    
    x = 10x - 10
    
    9x = 10
    
    x = 1.111111
    
    Check:
    
    log 1.111111 - log 0.111111 = 
    0.0457574906 + 0.9542425094 = 1
    

18. Solve by changing the base: x = lg 17

                ln 17    2.8332133440
    x = lg 17 = ------ = ------------ = 4.08746284
                ln 2     0.6931471806
    

19. Solve by changing the base: x = log₅ 100

                   log 100      2.0
    x = log5 100 = -------- = -------- = 2.861353
                    log 5     0.69897
    

20. Calculate the energy released by a magnitude 6.4 earthquake in tons of TNT.

    Using:  E = 101.5×R + 4.8, we get:
    
    E = 101.5×6.4 + 4.8 = 1014.4 = 2.511886 × 1014 Joules
    
    Using:  1 ton of TNT = 4.184×109 Joules, we get: 
    
          2.511886 × 1014 [Joules]
    E = ----------------------------- = 60,035 Tons TNT
         4.184 × 109 [Joules/Ton TNT]