1. 4.4 Exponential &
Logarithmic Equations
Day 2
Romans 15:4 For whatever was written in former
days was written for our instruction, that through
endurance and through the encouragement of the
Scriptures we might have hope.

2. Solve:
1. 4 2 x−1
− 27 = 0

3. Solve:
1. 4 2 x−1
− 27 = 0
2 x−1
4 = 27

4. Solve:
1. 4 2 x−1
− 27 = 0
2 x−1
4 = 27
2 x−1
log 4 = log 27

5. Solve:
1. 4 2 x−1
− 27 = 0
2 x−1
4 = 27
2 x−1
log 4 = log 27
( 2x − 1) log 4 = log 27

6. Solve:
1. 4 2 x−1
− 27 = 0
2 x−1
4 = 27
2 x−1
log 4 = log 27
( 2x − 1) log 4 = log 27
log 27
2x − 1 =
log 4

7. Solve:
1. 4 2 x−1
− 27 = 0
2 x−1
4 = 27
2 x−1
log 4 = log 27
( 2x − 1) log 4 = log 27
log 27
2x − 1 =
log 4
log 27
2x = +1
log 4

8. Solve:
1. 4 2 x−1
− 27 = 0
2 x−1
4 = 27 log 27 1
log 4 2 x−1
= log 27 x= +
2 log 4 2
( 2x − 1) log 4 = log 27 this is now
log 27 “calculator ready”
2x − 1 =
log 4
log 27
2x = +1
log 4

9. Solve:
1. 4 2 x−1
− 27 = 0
2 x−1
4 = 27 log 27 1
log 4 2 x−1
= log 27 x= +
2 log 4 2
( 2x − 1) log 4 = log 27 this is now
log 27 “calculator ready”
2x − 1 =
log 4 x ≈ 1.6887
log 27
2x = +1
log 4

10. Solve:
2. e = x − 1
x 2

11. Solve:
2. e = x − 1
x 2
no algebraic method presents itself ...
do graphically

12. Solve:
2. e = x − 1
x 2
no algebraic method presents itself ...
do graphically
x 2
y1 = e y2 = x − 1

13. Solve:
2. e = x − 1
x 2
no algebraic method presents itself ...
do graphically
x 2
y1 = e y2 = x − 1
find intersection points

14. Solve:
2. e = x − 1
x 2
no algebraic method presents itself ...
do graphically
x 2
y1 = e y2 = x − 1
find intersection points
x ≈ −1.1478

15. Solve:
3. x > log x − 2

16. Solve:
3. x > log x − 2
one variable ... one axis

17. Solve:
3. x > log x − 2
one variable ... one axis
graph : x − log x + 2 > 0

18. Solve:
3. x > log x − 2
one variable ... one axis
graph : x − log x + 2 > 0
x > 0 or ( 0,∞ )

19. Solve:
4. y < log ( 2x ) + 1

20. Solve:
4. y < log ( 2x ) + 1
two variables ... two axes
graph on graph paper!

21. Solve:
4. y < log ( 2x ) + 1
two variables ... two axes
graph on graph paper!
graph : y1 = log ( 2x ) + 1
dotted line & shade below

22. Solve:
5. x
y≥e −2

23. Solve:
5. x
y≥e −2
two variables ... two axes
graph on graph paper!

24. Solve:
5. x
y≥e −2
two variables ... two axes
graph on graph paper!
x
graph : y1 = e − 2
solid line & shade above

25. Solve:
6. 5 3x−1
−7 4x
>0 (algebraically)

26. Solve:
6. 5 3x−1
−7 4x
>0 (algebraically)
3x−1 4x
5 >7

27. Solve:
6. 5 3x−1
−7 4x
>0 (algebraically)
3x−1 4x
5 >7
( 3x − 1) log 5 > 4x log 7

28. Solve:
6. 5 3x−1
−7 4x
>0 (algebraically)
3x−1 4x
5 >7
( 3x − 1) log 5 > 4x log 7
3x log 5 − log 5 > 4x log 7

29. Solve:
6. 5 3x−1
−7 4x
>0 (algebraically)
3x−1 4x
5 >7
( 3x − 1) log 5 > 4x log 7
3x log 5 − log 5 > 4x log 7
3x log 5 − 4x log 7 > log 5

30. Solve:
6. 5 3x−1
−7 4x
>0 (algebraically)
3x−1 4x
5 >7
( 3x − 1) log 5 > 4x log 7
3x log 5 − log 5 > 4x log 7
3x log 5 − 4x log 7 > log 5
x ( 3log 5 − 4 log 7 ) > log 5

31. Solve:
6. 5 3x−1
−7 4x
>0 (algebraically)
3x−1 4x
5 >7
( 3x − 1) log 5 > 4x log 7
3x log 5 − log 5 > 4x log 7
3x log 5 − 4x log 7 > log 5
x ( 3log 5 − 4 log 7 ) > log 5
do ( 3log 5 − 4 log 7 ) on calc

32. Solve:
6. 5 3x−1
−7 4x
>0 (algebraically)
3x−1 4x
5 >7 log 5
x<
( 3x − 1) log 5 > 4x log 7 3log 5 − 4 log 7
3x log 5 − log 5 > 4x log 7
3x log 5 − 4x log 7 > log 5
x ( 3log 5 − 4 log 7 ) > log 5
do ( 3log 5 − 4 log 7 ) on calc

33. Solve:
6. 5 3x−1
−7 4x
>0 (algebraically)
3x−1 4x
5 >7 log 5
x<
( 3x − 1) log 5 > 4x log 7 3log 5 − 4 log 7
3x log 5 − log 5 > 4x log 7
x < −.5446
3x log 5 − 4x log 7 > log 5
x ( 3log 5 − 4 log 7 ) > log 5
do ( 3log 5 − 4 log 7 ) on calc

34. HW #7
It is time for us all to stand and cheer for the doer,
the achiever – the one who recognizes the challenge
and does something about it.
Vince Lombardi