1. 4.4 Exponential &
Logarithmic Equations
Romans 12:12 Rejoice in hope, be patient in
tribulation, be constant in prayer.

2. To solve an exponential equation:

3. To solve an exponential equation:
1. Isolate the exponential expression

4. To solve an exponential equation:
1. Isolate the exponential expression
2. Take the log of both sides

5. To solve an exponential equation:
1. Isolate the exponential expression
2. Take the log of both sides
3. Solve and verify all solutions with
substitution

6. Solve:
1. 3 5 x+1
+ 6 = 10

7. Solve:
1. 5 x+1
3 + 6 = 10
5 x+1
3 =4

8. Solve:
1. 5 x+1
3 + 6 = 10
5 x+1
3 =4
5 x+1
log 3 = log 4

9. Solve:
1. 5 x+1
3 + 6 = 10
5 x+1
3 =4
5 x+1
log 3 = log 4
( 5x + 1) log 3 = log 4

10. Solve:
1. 5 x+1
3 + 6 = 10
5 x+1
3 =4
5 x+1
log 3 = log 4
( 5x + 1) log 3 = log 4
log 4
5x + 1 =
log 3

11. Solve:
1. 5 x+1
3 + 6 = 10
5 x+1
3 =4
5 x+1
log 3 = log 4
( 5x + 1) log 3 = log 4
log 4
5x + 1 =
log 3
log 4
5x = −1
log 3

12. Solve:
1. 5 x+1
3 + 6 = 10
5 x+1
3 =4 log 4 1
5 x+1 x= −
log 3 = log 4 5 log 3 5
( 5x + 1) log 3 = log 4
log 4
5x + 1 =
log 3
log 4
5x = −1
log 3

13. Solve:
1. 5 x+1
3 + 6 = 10
5 x+1
3 =4 log 4 1
5 x+1 x= −
log 3 = log 4 5 log 3 5
( 5x + 1) log 3 = log 4 x ≈ .05
log 4 and verify
5x + 1 =
log 3
log 4
5x = −1
log 3

14. Solve:
2. 8e = 24
2x

15. Solve:
2. 8e = 24
2x
2x
e =3

16. Solve:
2. 8e = 24
2x
2x
e =3
2x
ln e = ln 3

17. Solve:
2. 8e = 24
2x
2x
e =3
2x
ln e = ln 3
2x ln e = ln 3

18. Solve:
2. 8e = 24
2x
2x
e =3
2x
ln e = ln 3
2x ln e = ln 3
2x = ln 3

19. Solve:
2. 8e = 24
2x
2x
e =3
2x
ln e = ln 3
2x ln e = ln 3
2x = ln 3
ln 3
x=
2

20. Solve:
2. 8e = 24
2x
2x
e =3
2x
ln e = ln 3
2x ln e = ln 3
2x = ln 3
ln 3
x=
2
x ≈ .5493

21. Solve:
2. 8e = 24
2x
2x
e =3 could also do this
2x
ln e = ln 3 graphically
2x ln e = ln 3
2x = ln 3
ln 3
x=
2
x ≈ .5493

22. Solve:
2. 8e = 24
2x
2x
e =3 could also do this
2x
ln e = ln 3 graphically
2x ln e = ln 3 y1 = 8e2x
2x = ln 3 y2 = 24
ln 3
x=
2
x ≈ .5493

23. Solve:
2. 8e = 24
2x
2x
e =3 could also do this
2x
ln e = ln 3 graphically
2x ln e = ln 3 2x
y1 = 8e
2x = ln 3 y2 = 24
ln 3
x= find intersection point
2
x ≈ .5493

24. Solve:
3. 2x x
e −e −2=0

25. Solve:
3.
x
2x x
e −e −2=0 a quadratic in e

26. Solve:
3.
x
2x x
e −e −2=0 a quadratic in e
x 2 x 1
(e ) − (e ) − 2 = 0

27. Solve:
3.
x
2x x
e −e −2=0 a quadratic in e
x 2 x 1
(e ) − (e ) − 2 = 0
behaves like
2
x −x−2=0
( x − 2 )( x + 1) = 0

28. Solve:
3.
x
2x
e −e −2=0 x
a quadratic in e
(e x
− 2 ) ( e + 1) = 0
x x 2 x 1
(e ) − (e ) − 2 = 0
behaves like
2
x −x−2=0
( x − 2 )( x + 1) = 0

29. Solve:
3.
x
2x
e −e −2=0 x
a quadratic in e
(e x
− 2 ) ( e + 1) = 0
x x 2 x 1
(e ) − (e ) − 2 = 0
behaves like
x x
e −2=0 e +1 = 0
2
x −x−2=0
( x − 2 )( x + 1) = 0

30. Solve:
3.
x
2x
e −e −2=0 x
a quadratic in e
(e x
− 2 ) ( e + 1) = 0
x x 2 x 1
(e ) − (e ) − 2 = 0
behaves like
x x
e −2=0 e +1 = 0
x
e = −1 2
x −x−2=0
( x − 2 )( x + 1) = 0

31. Solve:
3.
x
2x
e −e −2=0 x
a quadratic in e
(e x
− 2 ) ( e + 1) = 0
x x 2 x 1
(e ) − (e ) − 2 = 0
behaves like
x x
e −2=0 e +1 = 0
x
e = −1 2
x −x−2=0
∅ ( x − 2 )( x + 1) = 0

32. Solve:
3.
x
2x
e −e −2=0 x
a quadratic in e
(e x
− 2 ) ( e + 1) = 0
x x 2 x 1
(e ) − (e ) − 2 = 0
behaves like
x x
e −2=0 e +1 = 0
x x
e =2 e = −1 2
x −x−2=0
∅ ( x − 2 )( x + 1) = 0

33. Solve:
3.
x
2x
e −e −2=0 x
a quadratic in e
(e x
− 2 ) ( e + 1) = 0
x x 2 x 1
(e ) − (e ) − 2 = 0
behaves like
x x
e −2=0 e +1 = 0
x x
e =2 e = −1 2
x −x−2=0
x
ln e = ln 2 ∅ ( x − 2 )( x + 1) = 0

34. Solve:
3.
x
2x
e −e −2=0 x
a quadratic in e
(e x
− 2 ) ( e + 1) = 0
x x 2 x 1
(e ) − (e ) − 2 = 0
behaves like
x x
e −2=0 e +1 = 0
x x
e =2 e = −1 2
x −x−2=0
x
ln e = ln 2 ∅ ( x − 2 )( x + 1) = 0
x ln e = ln 2

35. Solve:
3.
x
2x
e −e −2=0 x
a quadratic in e
(e x
− 2 ) ( e + 1) = 0
x x 2 x 1
(e ) − (e ) − 2 = 0
behaves like
x x
e −2=0 e +1 = 0
x x
e =2 e = −1 2
x −x−2=0
x
ln e = ln 2 ∅ ( x − 2 )( x + 1) = 0
x ln e = ln 2
x = ln 2

36. Solve:
4. 2 3x
x e = 4e 3x

37. Solve:
4. 2 3x
x e = 4e 3x
we could graph ... a good
idea here, but there is an
algebraic solution

38. Solve:
4. 2 3x
x e = 4e 3x
we could graph ... a good
idea here, but there is an
2 3x 3x
x e − 4e = 0 algebraic solution

39. Solve:
4. 2 3x
x e = 4e 3x
we could graph ... a good
idea here, but there is an
2 3x 3x
x e − 4e = 0 algebraic solution
e 3x
(x 2
− 4) = 0

40. Solve:
4. 2 3x
x e = 4e 3x
we could graph ... a good
idea here, but there is an
2 3x 3x
x e − 4e = 0 algebraic solution
e 3x
(x 2
− 4) = 0
3x 2
e =0 x −4=0

41. Solve:
4. 2 3x
x e = 4e 3x
we could graph ... a good
idea here, but there is an
2 3x 3x
x e − 4e = 0 algebraic solution
e 3x
(x 2
− 4) = 0
3x 2
e =0 x −4=0
∅

42. Solve:
4. 2 3x
x e = 4e 3x
we could graph ... a good
idea here, but there is an
2 3x 3x
x e − 4e = 0 algebraic solution
e 3x
(x 2
− 4) = 0
3x 2
e =0 x −4=0
2
∅ x =4

43. Solve:
4. 2 3x
x e = 4e 3x
we could graph ... a good
idea here, but there is an
2 3x 3x
x e − 4e = 0 algebraic solution
e 3x
(x 2
− 4) = 0
3x 2
e =0 x −4=0
2
∅ x =4
x = ±2

44. To solve a logarithmic equation:

45. To solve a logarithmic equation:
1. Isolate the log

46. To solve a logarithmic equation:
1. Isolate the log
2. Rewrite into exponential form

47. To solve a logarithmic equation:
1. Isolate the log
2. Rewrite into exponential form
3. Solve and verify

48. Solve:
1. log x = 4.2274

49. Solve:
1. log x = 4.2274
4.2274
10 =x

50. Solve:
1. log x = 4.2274
4.2274
10 =x
x ≈ 16881

51. Solve:
2. ln x − 3 = 5.7213

52. Solve:
2. ln x − 3 = 5.7213
ln x = 8.7213

53. Solve:
2. ln x − 3 = 5.7213
ln x = 8.7213
8.7213
e =x

54. Solve:
2. ln x − 3 = 5.7213
ln x = 8.7213
8.7213
e =x
x ≈ 6132.1

55. Solve:
3. log x 36 = 2

56. Solve:
3. log x 36 = 2
2
x = 36

57. Solve:
3. log x 36 = 2
2
x = 36
x=6

58. Solve:
3. log x 36 = 2
2
x = 36
x=6
recall ... the domain of logs states
{x : x > 0}

59. Solve:
4. log 2 ( 22 − x ) = 3

60. Solve:
4. log 2 ( 22 − x ) = 3
3
2 = 22 − x

61. Solve:
4. log 2 ( 22 − x ) = 3
3
2 = 22 − x
8 = 22 − x

62. Solve:
4. log 2 ( 22 − x ) = 3
3
2 = 22 − x
8 = 22 − x
x = 14

63. Solve:
5. 5 + 4 log ( 6x ) = 25

64. Solve:
5. 5 + 4 log ( 6x ) = 25
4 log ( 6x ) = 20

65. Solve:
5. 5 + 4 log ( 6x ) = 25
4 log ( 6x ) = 20
log ( 6x ) = 5

66. Solve:
5. 5 + 4 log ( 6x ) = 25
4 log ( 6x ) = 20
log ( 6x ) = 5
5
10 = 6x

67. Solve:
5. 5 + 4 log ( 6x ) = 25
4 log ( 6x ) = 20
log ( 6x ) = 5
5
10 = 6x
2
x = 16,666
3

68. Solve:
6. log ( x + 8 ) + log ( x − 1) = 1

69. Solve:
6. log ( x + 8 ) + log ( x − 1) = 1
can’t isolate ... could do graphically ...
but here is another tactic

70. Solve:
6. log ( x + 8 ) + log ( x − 1) = 1

71. Solve:
6. log ( x + 8 ) + log ( x − 1) = 1
log ⎡( x + 8 ) ( x − 1) ⎤ = 1
⎣ ⎦

72. Solve:
6. log ( x + 8 ) + log ( x − 1) = 1
log ⎡( x + 8 ) ( x − 1) ⎤ = 1
⎣ ⎦
log ( x + 7x − 8 ) = 1
2

73. Solve:
6. log ( x + 8 ) + log ( x − 1) = 1
log ⎡( x + 8 ) ( x − 1) ⎤ = 1
⎣ ⎦
log ( x + 7x − 8 ) = 1
2
1 2
10 = x + 7x − 8

74. Solve:
6. log ( x + 8 ) + log ( x − 1) = 1
log ⎡( x + 8 ) ( x − 1) ⎤ = 1
⎣ ⎦
log ( x + 7x − 8 ) = 1
2
1 2
10 = x + 7x − 8
2
x + 7x − 18 = 0

75. Solve:
6. log ( x + 8 ) + log ( x − 1) = 1
log ⎡( x + 8 ) ( x − 1) ⎤ = 1
⎣ ⎦
log ( x + 7x − 8 ) = 1
2
1 2
10 = x + 7x − 8
2
x + 7x − 18 = 0 QF

76. Solve:
6. log ( x + 8 ) + log ( x − 1) = 1
log ⎡( x + 8 ) ( x − 1) ⎤ = 1
⎣ ⎦
log ( x + 7x − 8 ) = 1
2
1 2
10 = x + 7x − 8
2
x + 7x − 18 = 0 QF
x = 2, − 9

77. Solve:
6. log ( x + 8 ) + log ( x − 1) = 1
log ⎡( x + 8 ) ( x − 1) ⎤ = 1
⎣ ⎦
log ( x + 7x − 8 ) = 1
2
1 2
10 = x + 7x − 8
2
x + 7x − 18 = 0 QF
x = 2, − 9 reject −9

78. Solve:
6. log ( x + 8 ) + log ( x − 1) = 1
log ⎡( x + 8 ) ( x − 1) ⎤ = 1
⎣ ⎦
log ( x + 7x − 8 ) = 1
2
1 2
10 = x + 7x − 8
2
x + 7x − 18 = 0 QF
x = 2, − 9 reject −9
x=2

79. Remember ... all of these can be done
graphically ... don’t forget that.
But you need to know how to do these
algebraically, if it is possible.
HW #7

80. HW #7
Excellence is not a singular act but a habit.
You are what you do repeatedly.
Shaquille O’Neal