Honors Precalculus:  Chapter 7 Proofs
Mt Lebanon HS 2004-5
David Kosbie

Note:  These proofs are "semi-formal".   Some steps are combined or elided for clarity or brevity.

Also note:  Due to editor limitations, subscripts in exponents are written as x_1 or log_b rather than x1 or logb.

Final note:  following each proof, there is a brief sketch on how to apply that proof.  This is one of several such applications.

1. Proof of the Add-Add Property of Linear Functions
2. Proof of the Multiply-Multiply Property of Power Functions
3. Proof of the Add-Multiply Property of Exponential Functions
4. Proof of the Constant-Second-Differences Property of Quadratic Functions
5. Proof of the Logarithm of a Power Property
6. Proof of the Logarithm of a Product Property
7. Proof of the Logarithm of a Quotient Property
8. Proof of the Change-of-Base Property of Logarithms
9. Proof of the Multiply-Add Property of Logarithmic Functions

1.  Proof of the Add-Add Property of Linear Functions
    Given:  f(x) = ax + b    and     x2 = c + x1  (where c is a constant)
    Prove:  f(x2) = ac + f(x1)
    Note:  Think about this:  for all linear functions, when you add a constant value -- c -- to x, you add another constant value -- ac --  to f(x).

Step Reason
1. f(x) = ax + b Given
2. x2 = c + x1 Given
3. f(x1) = ax1 + b Substitute x1 for x in Step 1
4. f(x2) = ax2 + b Substitute x2 for x in Step 1
5.       = a(c + x1) + b Substitute (c + x1) for x2 (see step 2)
6.       = ac + ax1 + b Distribute multiplication:  x(y + z) = xy + xz
7.       = ac + f(x1) Substitute f(x1) for ax1 + b (see step 3)
QED  

How to apply the Add-Add Property of Linear Functions:

Given the following table, derive the function f(x):

x f(x)
2 10
4 16
5 19
6 22
  1. We first confirm the Add-Add Pattern.
  2. The points where x=2, x=4, and x=6 can be used with the constant c = 2
  3. Notice that the point where x=5 cannot be used, since this is not consistent with the pattern c=2.  Thus, we must skip this point for now and confirm that (5,19) lies on the line.  That is, we must confirm that f(5) = 19.
  4. 10 + 6 = 16 and 16 + 6 = 22, so indeed we are adding a constant, 6, to f(x) each time.
  5. Thus, we have confirmed the Add-Add pattern, where f(x1 + 2) = 6 + f(x1).  Thus, we know f(x) = ax + b.
  6. What's more, from the proof, we know f(x1 + 2) = ac + f(x1).  Thus, we have:
                f(x1 + 2) = 6 + f(x1)
    and
                f(x1 + 2) = ac + f(x1)
    thus:
                ac = 6
  7. Since ac=6 and c=2, we know that 2a=6, so a=3.
  8. Thus, we know that f(x) = 3x + b.  Now we must find b.  We do that by substituting the first point, (2,10):
                10 = (3)(2) + b
    so:
                b = 4
  9. Thus, we know that f(x) = 3x + 4.
  10. We must confirm the unused data points:  here, namely, (5,19):
                f(5) = (3)(5) + 4 = 19.  Check!
  11. Thus, f(x) = 3x + 4.

2.  Proof of the Multiply-Multiply Property of Power Functions
    Given:  f(x) = axb    and     x2 = cx1  (where c is a constant)
    Prove:  f(x2) = cb f(x1)
    Note:  Think about this:  for all power functions, when you multiply x by a constant value -- c -- you multiply f(x) by another constant value -- cb.

Step Reason
1. f(x) = axb Given
2. x2 = cx1 Given
3. f(x1) = ax1b Substitute x1 for x in Step 1
4. f(x2) = ax2b Substitute x2 for x in Step 1
5.       = a(cx1)b Substitute (cx1) for x2 (see step 2)
6.       = a(cbx1b) Distribute exponentiation:  (xy)z = xzyz
7.       = acbx1b Associate multiplication:  x(yz) = xyz
8.       = cbax1b Commute multiplication:  xyz = yxz
9.       = cb f(x1) Substitute f(x1) for ax1b (see step 3)
QED  

How to apply the Multiply-Multiply Property of Power Functions:

Given the following table, derive the function f(x):

x f(x)
1 3
2 24
3 81
4 192
  1. We first confirm the Multiply-Multiply Pattern.
  2. The points where x=1, x=2, and x=4 can be used with the constant c = 2
  3. Notice that the point where x=3 cannot be used, since this is not consistent with the pattern c=2.  Thus, we must skip this point for now and confirm that (3,81) lies on the power function's graph.  That is, we must confirm that f(3) = 81.
  4. (3)(8) = 24 and (24)(8) = 192, so indeed we are multiplying f(x) by a constant, 8, each time.
  5. Thus, we have confirmed the Multiply-Multiply pattern, where f(2x1) = 8 f(x1).  Thus, we know f(x) = axb.
  6. What's more, from the proof, we know f(2x1) = cb f(x1).  Thus, we have:
                f(2x1) = 8 f(x1)
    and
                f(2x1) = cb f(x1)
    thus:
                cb = 8
  7. Since cb = 8 and c=2, we know that 2b = 8, so b=3.
  8. Thus, we know that f(x) = ax3.  Now we must find a.  We do that by substituting the first point, (1,3):
                3 = a(1)3
    so:
                a= 3
  9. Thus, we know that f(x) = 3x3.
  10. We must confirm the unused data points:  here, namely, (3,81):
                f(3) = (3)(3)3 = 81.  Check!
  11. Thus, f(x) = 3x3.

3.  Proof of the Add-Multiply Property of Exponential Functions
    Given:  f(x) = abx    and     x2 = c + x1  (where c is a constant)
    Prove:  f(x2) = bc f(x1)
    Note:  Think about this:  for all exponential functions, when you add a constant value -- c -- to x, you multiply f(x) by another constant value -- bc.

Step Reason
1. f(x) = abx Given
2. x2 = c + x1 Given
3. f(x1) = abx_1 Substitute x1 for x in Step 1
4. f(x2) = abx_2 Substitute x2 for x in Step 1
5.       = ab(c + x_1) Substitute (c + x1) for x2 (see step 2)
6.       = a(bcbx_1) Distribute exponentiation:  x(y+z) = xyxz
7.       = abcbx_1 Associate multiplication:  x(yz) = xyz
8.       = bcabx_1 Commute multiplication:  xyz = yxz
9.       = bc f(x1) Substitute f(x1) for abx_1 (see step 3)
QED  

How to apply the Add-Multiply Property of Exponential Functions:

Given the following table, derive the function f(x):

x f(x)
1 10
3 40
4 80
5 160
  1. We first confirm the Multiply-Multiply Pattern.
  2. The points where x=3, x=4, and x=5 can be used with the constant c = 1 (or you could use x=1, x=3, and x=5 with c=2, either way will obtain the same answer, just be consistent throughout your solution.)
  3.  Notice that the point where x=1 cannot be used, since this is not consistent with the pattern c=1.  Thus, we must skip this point for now and confirm that (1,10) lies on the exponential function's graph.  That is, we must confirm that f(1) = 10.
  4. (40)(2) = 80 and (80)(2) = 160, so indeed we are multiplying f(x) by a constant, 2, each time.
  5. Thus, we have confirmed the Add-Multiply pattern, where f(1 + x1) = 2 f(x1).  Thus, we know f(x) = abx.
  6. What's more, from the proof, we know f(1 + x1) = bc f(x1).  Thus, we have:
                f(1 + x1) = 2 f(x1)
    and
                f(1 + x1) = bc f(x1)
    thus:
                bc = 2
  7. Since bc = 2 and c=1, we know that b1 = 2, so b=2.
  8. Thus, we know that f(x) = a2x.  Now we must find a.  We do that by substituting the first point we used, (3,40):
                40 = a(2)3
    so:
                a= 5
  9. Thus, we know that f(x) = 5(2)x.
  10. We must confirm the unused data points:  here, namely, (1,10):
                f(1) = 5(2)1 = 10.  Check!
  11. Thus, f(x) = 5(2)x.

4.  Proof of the Constant-Second-Differences Property of Quadratic Functions
    Given:  f(x) = ax2 + bx + c   and    x2 = d + x1    and  x3 = 2d + x1  (where d is a constant)
    Prove: [ f(x3) - f(x2) ]  - [ f(x2) - f(x1) ] = 2ad2.
    Note:  Think about this:  for all quadratic functions, when you add a constant value -- d -- repeatedly to x, then the second differences (the differences of the differences) of the y-values is always another constant -- 2ad2.
    Also Note:  this is a handy way to quickly find "a" in the equation -- find the constant second difference, set it equal to 2ad2, and solve for a.

Step Reason
1. f(x) = ax2 + bx + c Given
2. x2 = d + x1 Given
3. x3 = 2d + x1 Given
4. f(x1) = ax12 + bx1 + c Substitute x1 for x in Step 1

Subgoal:  Find f(x2) in terms of f(x1)

  
5. f(x2) = ax22 + bx2 + c Substitute x2 for x in Step 1
6.       = a(d + x1)2 + b(d + x1) + c Substitute (d + x1) for x2 (see step 2)
7.       = a(d2 + 2dx1 + x12) + b(d + x1) + c FOIL: (d + x1)2 = d2 + 2dx1 + x12
8.       = ad2 + 2adx1 + ax12 + b(d + x1) + c Distribute multiplication:  w(x + y + z) = wx + wy + wz
9.       = ad2 + 2adx1 + ax12 + bd + bx1 + c Distribute multiplication:  x(y + z) = xy + xz
10.      = ad2 + 2adx1 + bd + ax12 + bx1 + c Commute addition:  x + y + z = x + z + y
11.      = ad2 + 2adx1 + bd + f(x1) Substitute f(x1) for ax12 + bx1 + c (see step 4)

Subgoal:  Find f(x3) in terms of f(x1)

  
12. f(x3)= ax32 + bx3 + c Substitute x3 for x in Step 1
13.      = a(2d + x1)2 + b(2d + x1) + c Substitute (2d + x1) for x3 (see step 3)
14.      = a(4d2 + 4dx1 + x12) + b(2d + x1) + c FOIL: (2d + x1)2 = 4d2 + 4dx1 + x12
15.      = 4ad2 + 4adx1 + ax12 + b(2d + x1) + c Distribute multiplication:  w(x + y + z) = wx + wy + wz
16.      = 4ad2 + 4adx1 + ax12 + 2bd + bx1 + c Distribute multiplication:  x(y + z) = xy + xz
17.      = 4ad2 + 4adx1 + 2bd + ax12 + bx1 + c Commute addition:  x + y + z = x + z + y
18.      = 4ad2 + 4adx1 + 2bd + f(x1) Substitute f(x1) for ax12 + bx1 + c (see step 4)

Subgoal:  Find first difference f(x2) - f(x1)

  
19. f(x2) - f(x1)
         = ad2 + 2adx1 + bd + f(x1) - f(x1)		 Substitute ad2 + 2adx1 + bd + f(x1) for f(x2) (see step 11)
20.      = ad2 + 2adx1 + bd Subtraction:  f(x1) - f(x1) = 0

Subgoal:  Find first difference f(x3) - f(x2)

  
21. f(x3) - f(x2)
         = 4ad2 + 4adx1 + 2bd + f(x1) - f(x2)		 Substitute 4ad2 + 4adx1 + 2bd + f(x1) for f(x3) (see step 18)
22.      = 4ad2 + 4adx1 + 2bd + f(x1)
           - (ad2 + 2adx1 + bd + f(x1))		 Substitute ad2 + 2adx1 + bd + f(x1) for f(x2) (see step 11)
23.      = 4ad2 + 4adx1 + 2bd + f(x1)
           -ad2 - 2adx1 -  bd - f(x1)		 Distribute multiplication: -(w + x + y + z) = -w-x-y-z
24.      = 3ad2 + 2adx1 +  bd Add like terms

Subgoal:  Find second difference

  
25. [f(x3) - f(x2)] - [f(x2) - f(x1)]
         = (3ad2 + 2adx1 +  bd)
           - (ad2 + 2adx1 + bd)		 Substitute from steps 20 and 24
26.      = 3ad2 + 2adx1 +  bd
           -ad2 - 2adx1 -  bd		 Distribute multiplication: -(x + y + z) = -x-y-z
27.      = 2ad2 Add like terms
QED  

How to apply the Constant-Second-Differences Property of Quadratic Functions:

Given the following table, derive the function f(x):

x f(x)
1 2
3 22
5 66
6 97
7 134
  1. We first confirm the Constant-Second-Differences Pattern.
  2. The points where x=1, x=3, x=5, and x=7 can be used with the constant d = 2
  3.  Notice that the point where x=6 cannot be used, since this is not consistent with the pattern d=2.  Thus, we must skip this point for now and confirm that (6,97) lies on the quadratic function's graph.  That is, we must confirm that f(6) = 97.
  4. Consider the following table:
     
    x f(x) 1st
    diffs    		 2nd
    diffs
    1 2    
        20  
    3 22   24
        44  
    5 66   24
        68  
    7 134    

     

  5. Thus, we have confirmed the Constant-Second-Differences pattern, where d=2 and the constant second differences equal 24.
  6. What's more, from the proof, we know the constant second differences equal 2ad2, so:
                2ad2 = 24
    thus:
                2a(2)2 = 24
    so:
                a = 3
  7. Thus, we know f(x) = ax2 + bx + c and a=3, so f(x) = 3x2 + bx + c.  We now find b and c by substituting the first two points we used -- namely, (1,2) and (3,22):
                2 = 3(1)2 + b(1) + c
                22 = 3(3)2 + b(3) + c
  8. Subtracting the first equation from the second gives us:
                20 = 24 + 2b
    so:
                b = -2
  9. To find c, substitute b = -2 back into one of the previous equations, say 2 = 3(1)2 + b(1) + c, to get:
                2 = 3(1)2 + (-2)(1) + c
    so:
                c = 1
  10. Thus, we know that f(x) = 3x2 - 2x + 1.
  11. We must confirm the unused data points:  here, namely, (6,97):
                f(6) = 3(6)2 - 2(6) + 1 = 108 - 12 + 1 = 97.  Check!
  12. Thus, f(x) = 3x2 - 2x + 1.

5.  Proof of the Logarithm of a Power Property
    Prove:  logb(xy) = y logbx
    We will restate this as the following:
         Given:  logbc = y logbx
         Prove: c = xy

Step Reason
1. logbc = y logbx Given
2. b(log_b c) = b(y log_b(x)) If x = y then bx = by
3. c        = b(y log_b(x)) b(log_b c) = c
4.          = b(log_b(x) y) Commute Multiplication:  ab = ba
5.          = (b(log_b(x))y Power of Products Property:  a(bc) = (ab)c
6.          = xy b(log_b x) = x
QED  

Application:  Solve for x:  ln(2x) = 3.

Step Reason
1. ln(2x) = 3 Given
2. x ln 2 = 3 Log of a power property:  logb(xy) = y logbx
3. x = 3 / ln(2) Division
4.   = 4.328... Use a calculator

Check:  ln(24.328) = ln(20.0844) = 3.000.  Check!

6.  Proof of the Logarithm of a Product Property
    Prove:  logb(xy) = logbx + logby
    We will restate this as the following:
         Given:  logbc = logbx + logby
         Prove: c = xy

Step Reason
1. logbc = logbx + logby Given
2. b(log_b c) = b(log_b(x) + log_b(y)) If x = y then bx = by
3. c        = b(log_b(x) + log_b(y)) Defn of Logarithm:  b(log_b c) = c
4.          = blog_b(x)blog_b(y) Power of Sums Property:  a(b+c) = abac
5.          = x blog_b(y) Defn of Logarithm:  b(log_b x) = x
6.          = xy Defn of Logarithm:  b(log_b y) = y
QED  

Application:  Solve for x:  log(x2) + log(x-1) = 3.

Step Reason
1. log(x2) + log(x-1) = 3 Given
2. log(x2 * x-1) = 3 Log of a product property:  logb(xy) = logbx + logby
3. log(x) = 3 Power of Sums Property:  a(b+c) = abac
4. 10log(x) = 103 If x=y then 10x = 10y
5. x      =  103 Defn of Logarithm:  b(log_b x) = x, so 10(log x) = x
6.        = 1000 Arithmetic

Check:  log(10002) + log(1000-1) = 6 - 3 = 3.  Check!

7.  Proof of the Logarithm of a Quotient Property
    Prove:  logb(x/y) = logbx - logby
    We will restate this as the following:
         Given:  logbc = logbx - logby
         Prove: c = x/y

Step Reason
1. logbc = logbx - logby Given
2. b(log_b c) = b(log_b(x) - log_b(y)) If x = y then bx = by
3. c        = b(log_b(x) - log_b(y)) Defn of Logarithm:  b(log_b c) = c
4.          = blog_b(x) / blog_b(y) Power of Differences Property:  a(b-c) = ab / ac
5.          = x / blog_b(y) Defn of Logarithm:  b(log_b x) = x
6.          = x/y Defn of Logarithm:  b(log_b y) = y
QED  

Application:  Solve for x:  log(3/x) = 3.

Step Reason
1. log(3/x) = 3 Given
2. log(3) - log(x) = 3 Log of a quotient property:  logb(x/y) = logbx - logby
3. log(x) = log(3) - 3 Arithmetic
4.        = -2.5229... Use a calculator
5. 10log(x) = 10-2.5229 If x=y then 10x = 10y
6. x      = 10-2.5229 Defn of Logarithm:  b(log_b x) = x, so 10(log x) = x
7.        = 0.003 Use a calculator

Check:  log(3/0.003) = log(1000) = 3.  Check!

Note:  Actually, this example is a bit forced, since you could skip the log-of-quotient property as such:

Step Reason
1. log(3/x) = 3 Given
2. 10log(3/x) = 103 If x=y then 10x = 10y
3. 3/x      = 103 Defn of Logarithm:  b(log_b x) = x, so 10(log x) = x
4.          = 1000 Arithmetic.
5. x/3 = .001 If x=y then 1/x = 1/y
6. x = .003 Multiply.

8.  Proof of the Change-of-Base Property of Logarithms
    Prove:  logax = logbx / logba, or, equivalently:  logbx = (logba)(logax)
    We will restate this as the following:
         Given:  logbc = (logba)(logax)
         Prove: c = x

Step Reason
1. logbc = (logba)(logax) Given
2. b(log_b c) = b(log_b(a) log_a(x)) If x = y then bx = by
3. c        = b(log_b(a) log_a(x)) Defn of Logarithm:  b(log_b c) = c
4.          = (blog_b(a))log_a(x) Power of Products Property:  x(yz) = (xy)z
5.          = alog_a(x) Defn of Logarithm:  b(log_b a) = a
6.          = x Defn of Logarithm:  a(log_a x) = x
QED  

Application: Evaluate log35.
The problem is that your calculator does not have a log3x button.  So we must convert this into either base-10 or base-e.  We'll use base-e, just for fun.

Step Reason
1. log35 = loge5 / loge3 Change-of-base property:  logax = logbx / logba
2.       = ln(5) / ln(3) Notation:  ln(x) = logex
3.       = 1.6094 / 1.0986 Use a calculator
4.        = 1.4650 Ditto

Check:  We can check by raising 31.4650 (why?), which your calculator does support.  We get:
      31.4650 =  5
Check!

9. Proof of the Multiply-Add Property of Logarithmic Functions
    Given:  f(x) = a + blogcx   and     x2 = kx1  (where k is a constant)
    Prove:  f(x2) = blogck + f(x1)
    Note #1:  This is presented erroneously in the book (see p. 302), which omits the constant "b" in f(x2) = blogck + f(x1).
    Note #2:  Think about this:  for all exponential functions, when you multiply x by a constant value -- k -- you add another constant value -- blogck --  to f(x).

Step Reason
1. f(x) = a + blogcx Given
2. x2 = kx1 Given
3. f(x1) = a + blogcx1 Substitute x1 for x in Step 1
4. f(x2) = a + blogcx2 Substitute x2 for x in Step 1
5.       = a + blogc(kx1) Substitute kx1 for x2 (see step 2)
6.       = a + b(logck + logcx1) Log of a product (see proof above!): logb(xy) = logbx + logby
7.       = a + blogck + blogcx1 Distribute multiplication:  x(y + z) = xy + xz
8.       = blogck + a + blogcx1 Commute addition:  x + y + z = y + x + z
9.       = blogck + f(x1) Substitute f(x1) for (a + blogcx1)(see step 3)
QED  

How to apply the Multiply-Add Property of Logarithmic Functions:

Given the following table, derive the function f(x):

x f(x)
4 13
12 20.92
16 23
64 33
  1. We first confirm the Multiply-Add Pattern.
  2. The points where x=4, x=16, and x=64 can be used with the constant k = 4.
  3.  Notice that the point where x=12 cannot be used, since this is not consistent with the pattern k=4.  Thus, we must skip this point for now and confirm that (12,20.92) lies on the logarithmic function's graph.  That is, we must confirm that f(12) = 20.92.
  4. 13 + 10 = 23 and 23 + 10 = 33, so indeed we are adding a constant, 10, to f(x) each time.
  5. Thus, we have confirmed the Multiply-Add pattern, where f(4x1) = 10 + f(x1).  Thus, we know f(x) = a + blogcx.
  6. What's more, from the proof, we know f(4x1) = blogck + f(x1).  Thus, we have:
                f(4x1) = 10 + f(x1)
    and
                f(4x1) = blogck + f(x1)
    thus:
                blogck = 10
  7. Since blogck = 10 and k=4, we know that blogc4 = 10.
  8. We are now free to choose our base for our final answer.  We will choose c=2.
  9. Thus, blog24 = 10.  But log24 = 2, so 2b = 10, so b=5.
  10. Thus, we know that f(x) = a + 5log2x.  Now we must find a.  We do that by substituting the first point we used, (4,13):
                13 = a + 5log24
    so:
                a= 3
  11. Thus, we know that f(x) = 3 + 5log2x.
  12. We must confirm the unused data points:  here, namely, (12,20.92):
                f(12) = 3 + 5log2(12) = 3 + 5( ln(12) / ln(2) ) = 20.92...  Check!
  13. Thus, f(x) = 3 + 5log2x.