1. 5.6 Laws of Logarithms 5.7 Exponential Equations; Changing Base
Objectives:
1. Compare & recall the properties of exponents
2. Deduce the properties of logarithms from/by comparing the properties of exponents
Use the properties of logarithms
Solve the exponential equations

2. Since the logarithmic function y = logb x is the inverse of the exponential function y = bx, the laws of logarithms are very closely related to the laws of exponent.
Pre-Knowledge
For any b, c, u, v  +, and b ≠ 1, c ≠ 1, there exists some x, y  , such that
u = bx, v = by
By the previous section knowledge, as long as taking
x = logbu, y = logbv

3. logbuv = logb(bxby)= logbb x+y = x + y
1. Product of Power
am an = am+n
Pre-Knowledge
For any b, c, u, v  +, and b ≠ 1, c ≠ 1, there exists some x, y  , such that
u = bx, v = by
By the previous section knowledge, as long as taking
x = logbu, y = logbv
1. Product Property
logbuv = logbu + logbv
Proof
logbuv = logb(bxby)= logbb x+y = x + y
= logbu + logbv

4. 2. Quotient of Power 2. Quotient Property Proof Pre-Knowledge
For any b, c, u, v  +, and b ≠ 1, c ≠ 1, there exists some x, y  , such that
u = bx, v = by
By the previous section knowledge, as long as taking
x = logbu, y = logbv
2. Quotient Property
Proof

5. Equal Power am = an iff m = n Equal Property logbu = logbv iff u = v
Proof
logbu – logbv = 0

6. logbuk = logb(bx)k = logbb kx = kx = k logbu
4. Power of Power
(am)n = amn
Pre-Knowledge
For any b, c, u, v  +, and b ≠ 1, c ≠ 1, there exists some x, y  , such that
u = bx, v = by
By the previous section knowledge, as long as taking
x = logbu, y = logbv
4. Power Property
logbuk = k logbu
Proof
logbuk = logb(bx)k = logbb kx = kx = k logbu

7. 5. Change-of-Base Formula
Proof Note that
bx = u, logbu = x
Taking the logarithm with base c at both sides:
logcbx = logcu or x logcb = logcu

8. 6. Reciprocal Formula
Proof

9. 7. Raise Power Formula
Proof

10. Example 1 Assume that log95 = a, log911 = b, evaluate

11. Example 2 Expanding the expression
ln(3y4/x3)
ln(3y4/x3) = ln(3y4) – lnx3 = ln3 + lny4 – lnx3
= ln3 + 4 ln|y| – 3 lnx
b) log3125/6x9
log3125/6x9 = log3125/6 + log3x9
= 5/6 log log3x
= 5/6 log3(3· 22) + 9 log3x
= 5/6 (log33 + log322) + 9 log3x
= 5/6 ( log32) + 9 log3x

12. Example 3 Condensing the expression
a) 3 ( ln3 – lnx ) + ( lnx – ln9 )
3 ( ln3 – lnx ) + ( lnx – ln9 )
= 3 ln3 – 3 lnx + lnx – 2 ln3
= ln3 – 2 lnx
= ln(3/x2)
b) 2 log37 – 5 log3 x + 6 log9 y2
2 log37 – 5 log3 x + 6 log9 y2
= log349 – log3 x5 + 6 ( log3 y2/ log39)
= log3(49/x5) + 3 log3 y2
= log3(49y6/x5)

13. True or False log a y = – log 1/a y
Practice
A) P. 199 Q 7 – 18
P. 199 Q 19 – 20 How do you change to make it to be true?
True or False log a y = – log 1/a y
c) P. 200 Q 7 – 27 (odd)

14. Example 4 Calculate log48 and log615 using common and natural logarithms.
a) log48
log48 = log8 / log4 = 3 log2 / (2 log2)
= 3/2
log48 = ln8 / ln4 = 3 ln2 / (2 ln2) = 3/2
b) log615 = log15 / log6 = 1.511

15. in terms of
Example 5. Express

16. More on Expand/Condense logarithmic expressions
Example 6 Expand

17. More on Expand/Condense logarithmic expressions
Example 7 Expand in terms of sums and differences of logarithms

18. More on Expand/Condense logarithmic expressions
Example 8 Expand to express all powers as factors

19. More on Expand/Condense logarithmic expressions
Example 9 Condense to a single logarithm.

20. Assignment:
5.6
P #36 – 44 (even)
P #2 – 22 (even), 21 – 33 (odd), 41, 43, 45

21. Solving Exponential Equations
One way to solve exponential equations is to use the property that if 2 powers with the same base are equal, then their exponents are equal.
For b > 0 and b≠1
if bx = by, then x = y

22. Solve by Equating Exponents
Example 10: Solve 43x = 8x+1
(22)3x = (23)x+1 rewrite with same base
26x = 23x+3
6x = 3x + 3
x = 1
Check → 43*1 = 81+1
64 = 64

23. 24x = 32x–1 Solve: 16x = 32x–1 Your turn! 24x = (25)x–1 4x = 5x – 5
Be sure to check your answer!!!

24. When you can’t rewrite using the same base, you can solve by taking a log of both sides or use the definition of log
Example 11: Solve 2x = 7
log22x = log27
x = log27
x = ≈ 2.807

25. 102x – 3 = 17 log10102x – 3 = log1017 2x – 3 = log 17 2x = 3 + log17
When you can’t rewrite using the same base, you can solve by taking a log of both sides or use the definition of log
Example 12: Solve 102x – = 21
102x – 3 = 17
log10102x – 3 = log1017
2x – 3 = log 17
2x = 3 + log17
x = ½(3 + log17)
≈ 2.115

26. Your turn! Solve: 5x + 2 + 3 = 25 5x+2 = 22 log55x+2 = log522
= (log22/log5) – 2
≈ –0.079

27. More on Solving Exponential Equations
Example 13: Solve ex – 3e-x = 2
[Answer] Multiply ex at both sides of the equation:
ex(ex – 3e-x ) = 2ex
e2x – 3 = 2ex
e2x – 2ex – 3 = 0
(ex)2 – 2(ex) – 3 = 0
Denote ex = u, then
(u)2 – 2(u) – 3 = 0
(u)2 – 2(u) – 3 = 0
(u – 3)(u + 1) = 0
u = 3, or u = –1
ex = 3, or ex = –1 (discard)
x = ln3

28. Solving Logarithmic Equations
1. To solve use the property for logs with the same base:
b, x, y+ and b  1
If logb x = logb y, then x = y
2. When you can’t rewrite both sides as logs with the same base exponentiate each side
b, x+ and b  1
if logb x = y, then x = by
This can get the expression in the log out of the log simply.

29. Example 14: Newton’s Law of Cooling
Solving Logarithmic Equations
Example 14: Newton’s Law of Cooling
The temperature T of a cooling substance at time t (in minutes) is:
T = (T0 – TR) e-rt + TR
T0= initial temperature
TR= room temperature
r = constant cooling rate of the substance

30. Solving Logarithmic Equations
Example 14: You’re cooking stew. When you take it off the stove the temp. is 212°F. The room temp. is 70°F and the cooling rate of the stew is r = How long will it take to cool the stew to a serving temp. of 100°?
T = (T0 – TR) e -rt + TR
T0 = 212, TR = 70, T = r = 0.046
So solve:
100 = (212 – 70)e t + 70

31. Solving Logarithmic Equations
30 = 142e t (subtract 70)
15/71 = e t (divide by 142)
How do you get the variable out of the exponent?
ln(15/71) = lne-.046t (take the ln of both sides)
ln(15/71) = – 0.046t
ln(15/71)/(– 0.046) = t
t = (ln15 – ln71)/(– 0.046) =
t ≈ – 1.556/(– 0.046)
t ≈ about 34 minutes to cool!

32. Example 15: Solve log3(5x – 1) = log3(x + 7)
Solving Logarithmic Equations
Example 15: Solve log3(5x – 1) = log3(x + 7)
5x – 1 = x + 7
4x = 8
x = and check
log3(52 – 1) = log3(2 + 7)
log39 = log39
Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions.

33. Example 16: Solve log5(3x + 1) = –2
Solving Logarithmic Equations
Example 16: Solve log5(3x + 1) = –2
3x + 1 = 5-2
3x + 1 = 1/25
x = –8/25 and check
Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions.

34. Example 16: Solve log5x + log(x + 1) = 2
More on Solving Logarithmic Equations
Example 16: Solve log5x + log(x + 1) = 2
log[5x(x + 1)]= (product property)
log (5x2 + 5x) = 2
5x2 + 5x = 100
x2 + x – 20 = (subtract 100 and divide by 5)
(x + 5)(x – 4) = x = – 5, or x = 4
check and you’ll see x = 4 is the only solution.

35. Your Turn! Solve log2x + log2(x – 7) = 3
log2 [x(x – 7)]= 3
log2 (x2 – 7x) = 3
x2 – 7x = 23
x2 – 7x = 8
x2 – 7x – 8 = 0
(x – 8)(x + 1) = 0
x = 8 x = –1
Check
log28 + log2(8 – 7) =3
3 + 0 = 3

36. One More! Solve log6(x – 2) + log6(x + 3) = 2
log6 [(x – 2)(x + 3)] = 2
log6 (x2 + x – 6) = 2
x2 + x – 6 = 36
x2 + x – 42 = 0
(x – 6)(x + 7)=0
x = 6 x = –7
Check
log64 + log69 =2
log636 = 2
Check
log64 + log69 =2
log636 = 2

37. Challenge! Example 17: Solve log2x + log4(x2 – 4x + 4) = 3
log4 x2 + log4(x – 2)2 = 3 (Raise Power Formula)
log4 [x2(x – 2)2] = 3
x2(x – 2)2 = 43
[x(x – 2)]2 = 64 A2B2 = (AB)2
x(x – 2) = ±8 A2 = 64, A = ±8
x2 – 2x – 8 = or x2 – 2x + 8 =0
x = 4 x = –2 No real solution

38. Challenge! Solve log2x + ½ log2(x – 1) = 1
log2 [x2(x – 1)] = 2
x2(x – 1) = 4
x3– x2 – 4 = 0 (Rational Zero Theorem)
(x – 2)(x2 + x + 2) = 0
x – 2 = or x2 + x + 2 = 0
x = No real solution
Check
log22 + ½ log21 =1
2 + 0 = 2

39. Challenge Simplify (No calculator)1) 2) 3) 4)
5) Proof

40. Assignment:
5.7
P #24 – 34 (even), 42, 44
P #49 – 52