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.

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

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

 x f(x) 2 10 4 16 5 19 6 22
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.

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.

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.

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:
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!

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.