Chapter 12 - Algebra Coach Exercises

1. x^(1/5) * x^(2/5)
2. x^(3/5) / x^(1/5)
3. (x^(3/5))^2
4. x^1.2 * x^3
5. (x^1.2)^3
6. (x^5)^(-0.4)
7. (x^(a/b))^(c/d)
   
   
8. x^a * x^(c/d)
9. x^(a/c) * x^(b/c)
10. x^(a/b) / x^(c/d)

1. log(10^(-3))
2. log(b^x)
3. log(b^10 * c^15)
4. ln(e^5)
5. log(b^(3m) * c^(4m))
6. ln(b^x)
7. log(32)/log(8)

8. log(5) + log(8)
9. log(5) + log(2)
10. log(x) + log(y) + log(z)
11. 2*log(x) + 3*log(y) + z
12. a*log(x) - b*log(y) + 2*log(w+z)
13. log(x)/a + log(y)/b
14. log(x+2) + log(x-2)

1. y = 5 · 6 ^t       solve       6 ^ t = e ^ (r t)
2. y = 2 · 5 ^t/3       solve       5 ^ (t / 3) = e ^ (r t)

3. y = 4 · 6 ^− t       solve       6 ^ (-t) = e ^ (-r t)
4. y = 25 · (¼) ^− t       solve       (1 / 4) ^ (-t) = e ^ (-r t)

1. The graph of a certain exponential (growth) function contains the points (2, 3) and (6, 9). Find the equation of this exponential function in the rate form,  y = a e ^b x, where the constants a and b are to be determined. Here’s how:
   
   
   • Substitute the first point into the rate form. To do this drag-and-drop the following text into the Algebra Coach’s listbox and click the Substitute button:

     x = 2, y = 3, y = a * e ^ (b x)

     The result will be an equation containing a and b. Set the Autosolve variable to a and solve this equation for a. We will need to locate this solution later so type the words “Equation 1” beside it or near it.
     
     
   • Now repeat the whole procedure with the second point: Clear the Algebra Coach’s listbox and drag-and-drop the following text into it and click the Substitute button:

     x = 6, y = 9, y = a * e ^ (b x)

     The result will be a second equation containing a and b. Solve this equation for a and type the words “Equation 2” near it.
     
     
   • Clear the Algebra Coach’s listbox and drag both Equation 1 and Equation 2 (one after the other) into it. The listbox should now contain:

     a = 3 / e ^ (2*b), a = 9 / e ^ (6*b)

     Set the two expressions for a equal to each other by clicking the Substitute button. Set the Autosolve variable to b, set the exact or floating point option to floating point and solve this equation for b. The result will be b = 0.275. The Algebra Coach automatically checks this solution by substituting it back into the original equation. Since the original equation essentially reads a = a, this means that a = 1.73, and that the equation of the exponential function is:

     y = 1.73 e ^0.275 x

2. The graph of a certain exponential (decay) function contains the points (5, 22.7) and (15, 9.14). Find the equation of this exponential function in the rate form,  y = a e ^− b x, where the constants a and b are to be determined. Here’s how:
   
   
   • Substitute the first point into the rate form. To do this drag-and-drop the following text into the Algebra Coach’s listbox and click the Substitute button:

     x = 5, y = 22.7, y = a * e ^ (- b x)

     The result will be an equation containing a and b. Set the Autosolve variable to a and solve this equation for a. We will need to locate this solution later so type the words “Equation 1” beside it or near it.
     
     
   • Now repeat the whole procedure with the second point: Clear the Algebra Coach’s listbox and drag-and-drop the following text into it and click the Substitute button:

     x = 15, y = 9.14, y = a * e ^ (- b x)

     The result will be a second equation containing a and b. Solve this equation for a and type the words “Equation 2” near it.
     
     
   • Clear the Algebra Coach’s listbox and drag both Equation 1 and Equation 2 (one after the other) into it. The listbox should now contain:

     a = 22.7 * e ^ (5*b), a = 9.14 * e ^ (15*b)

     Set the two expressions for a equal to each other by clicking the Substitute button. Set the Autosolve variable to b and solve this equation for b. The result will be b = 0.091. The Algebra Coach automatically checks this solution by substituting it back into the original equation. Since the original equation essentially reads a = a, this means that a = 35.8, and that the equation of the exponential function is:

     y = 35.8 e ^{− 0.091 x}

1. Find the equation of the straight line in the graph to the right.
   
   First, note that it must be of the form i = a e ^b t, where a and b are to be determined, and that it contains the points (t = 2, i = 4) and (t = 27, i = 1).
   
   Drag-and-drop the following text into the Algebra Coach’s listbox, click the Substitute button, and then Autosolve for a:

   t = 2, i = 4, i = a e^(b t)

   Now do the same for the following text:

   t = 27, i = 1, i = a e^(b t)

   Now drag the two equations that were solved for a into the textbox. The textbox should now contain:

   a = 4 / e^(2b), a = 1 / e^(27b)

   Click the Substitute button, and then Autosolve for b: The result will be b = −0.0555. The Algebra Coach automatically checks this solution by substituting it back into the original equation. Since the original equation essentially reads a = a, this means that a = 4.47, and that the equation of the straight line is:

   i = 4.47 e ^{− 0.0555 t}

   
   
2. Find the equation of the straight line in the graph to the right.
   Follow the same steps as in the previous example. Note that the equation must be of the form P = a e ^b t, where a and b are to be determined, and that it contains the points (t = 4.5, P = 10) and (t = 6.6, P = 60).
   
   Therefore the relevent equations are:

   t = 4.5, P = 10, P = a e^(b t)
   t = 6.6, P = 60, P = a e^(b t)

   and the solution is:

   P = 0.215 e ^0.853 t

   
   
3. Find the equation of the straight line in the graph to the right.
   Follow the same steps as in the previous example. Note that the equation must be of the form C = a A ^b, where a and b are to be determined, and that it contains the points (A = 4.5, C = 10) and (A = 6.6, C = 60).
   
   Therefore the relevent equations are:

   A = 2, C = 1, C = a A^b
   A = 30, C = 9, C = a A^b

   and the solution is:

   C = 0.570 A ^0.811

   
   
4. Find the equation of the straight line in the graph to the right.
   Follow the same steps as in the previous example. Note that the equation must be of the form C = a A ^b, where a and b are to be determined, and that it contains the points (A = 4.5, C = 10) and (A = 6.6, C = 60).
   
   Therefore the relevent equations are:

   L = 0.1, Q = 10, Q = a L^b
   L = 3, Q = 1, Q = a L^b

   and the solution is:

   Q = 2.10 L ^− 0.677

   
   

1. 9 ^ x = 27
2. 2.24 ^ (x + 2) = 12.5
3. 4.12 e ^ (3 x) = 114
4. 3.26 ^ x = 86.8
5. e ^ (2 x) = 125
6. 1.05 e ^ (4 x + 1) = 5.96
   
   
7. e ^ (2 x - 1) = 3 e ^ (x + 3)
8. 5 ^ (2 x) = 8 ^ (3 x - 2)
9. 10 ^ (4 x) = 4 * 10 ^ x
10. 2 ^ (5 x + 1) = 3 ^ (2 x + 1)
11. 7 e ^ (1.5 x) = 2 e ^ (2.4 x)
12. 5 ^ (2 x) = 3 ^ (3 x + 1)
    
    
13. 3 ^ (x ^ 2) = 175 ^ (x - 1)
14. e ^ x + e ^ (-x) = 5
15. e ^ x + e ^ (-x) = 2 (e ^ x - e ^ (-x))
16. e ^ (4 x) - 2 e ^ (2 x) - 3 = 0
17. e ^ (6 x) - e ^ (3 x) - 2 = 0
18. e ^ (6 x) - 3 e ^ (3 x) + 2 = 0

1. log(2 x + 25) = 2
2. log(3 x + x ^ 2) = 1
3. 2log(x + 1)=3
4. log(8x) = log(24)
5. ln(8) + ln(x - 2) = ln(3x - 2)
   
   
6. ln(5x + 2) - ln(x + 6) = ln(4)
7. ln(x) + ln(x + 2) = 1
8. 2log(x) - log(1 - x) = 1
9. log(4x - 3) + log(5) = 6
10. ln(x) - 2ln(x) = ln(64)
    
    
11. ln(x + 2) - ln(36) = ln(x)
12. log(x) + log(4x) = 2
13. log(8x^2) - log(4x) = 2.54
14. 2log(x) - 1 = log(20 - 2x)
15. ln(2x) - ln(4) + ln(x - 2) = 1