The document is a lesson focusing on solving exponential equations and inequalities, detailing the laws of exponents and providing examples for clarity. It explains the differences between exponential equations and inequalities, listing various problems along with their solutions, emphasizing methods for both equality and inequality scenarios. The lesson also outlines two key rules for handling exponential inequalities and illustrates their application through example problems.

1. Lesson #5 :
Solving Exponential
Equations and Inequalities

2. Before anything else, let’s
recall the different
Laws of Exponent

3. Laws of Exponent
𝑎𝑚
. 𝑎𝑛
= 𝑎𝑚+𝑛
23. 24 = 23+4
𝑹𝒖𝒍𝒆 #𝟏
𝑹𝒖𝒍𝒆 #𝟐
(𝑎𝑚)𝑛 = 𝑎𝑚𝑛
(23)4 = 23(4)
𝑹𝒖𝒍𝒆 #𝟑
(𝑎𝑏)𝑚
= 𝑎𝑚
𝑏𝑚
(2.3)4 = 2434
𝑹𝒖𝒍𝒆 #𝟒
𝑎𝑚
𝑎𝑛
= 𝑎𝑚−𝑛, 𝑤ℎ𝑒𝑟𝑒 𝑎 ≠ 0
23
22
= 23−2
𝑹𝒖𝒍𝒆 #𝟔
𝑎0 = 1
20
= 1
𝑹𝒖𝒍𝒆 #𝟓
(
𝑎
𝑏
)𝑛 =
𝑎𝑛
𝑏𝑛
, 𝑤ℎ𝑒𝑟𝑒 𝑏 ≠ 0
(
2
3
)2 =
22
32

4. Laws of Exponent
𝑹𝒖𝒍𝒆 #𝟕 𝑹𝒖𝒍𝒆 #𝟖
𝑎−𝑥
=
1
𝑎𝑥
2−4
=
1
24
𝑎
𝑥
𝑦 =
𝑦
𝑎𝑥
2
3
4 =
4
23

5. Exponential Equation vs Inequality
Try This: Ask yourself this question: what’s the difference between equality and
inequality? Now, identify which item falls under exponential equation and
exponential inequality.
1) 10𝑥−2
= 1000010
2) 254+2𝑥 > 56𝑥
3)
1
16
10𝑥−3
≤ 4𝑥
4) 2𝑥2
= 32𝑥+3
5)
1
8
9−2𝑥
= 163
6) 124
𝑥
3 < 126

6. Solving Exponential Equation

7. Solving Exponential Equation
1) 2𝑥 = 4
𝑓𝑖𝑟𝑠𝑡 𝑠𝑡𝑒𝑝 𝑖𝑠 𝑡𝑜 𝑚𝑎𝑘𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑜𝑓 𝑏𝑜𝑡ℎ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑙
𝑖𝑛 𝑜𝑢𝑟 𝑒𝑥𝑎𝑚𝑝𝑙𝑒, 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝟐, 𝑚𝑎𝑘𝑒 4 𝑎𝑠 𝑎 𝑏𝑎𝑠𝑒 𝑜𝑓 2 (4 = 22)
2𝑥 = 22 𝑒𝑞𝑢𝑎𝑡𝑒 𝑡ℎ𝑒𝑖𝑟 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑠, 𝑡ℎ𝑒𝑛 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝒙
𝑥 = 2
𝐶ℎ𝑒𝑘𝑖𝑛𝑔: 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = 2 𝑡𝑜 𝑜𝑢𝑟 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
2𝑥
= 4
22 = 4
4 = 4
𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦
𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙, 𝑚𝑒𝑎𝑛𝑖𝑛𝑔 𝑜𝑢𝑟 𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡

8. Solving Exponential Equation
2) 5𝑥 =
1
5
𝑈𝑠𝑒 𝑅𝑢𝑙𝑒 #7 𝑖𝑛 𝐿𝑎𝑤𝑠 𝑜𝑓 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡
5𝑥= 5−1 𝑒𝑞𝑢𝑎𝑡𝑒 𝑡ℎ𝑒𝑖𝑟 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑠, 𝑡ℎ𝑒𝑛 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝒙
𝑥 = −1
𝐶ℎ𝑒𝑘𝑖𝑛𝑔: 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = 2 𝑡𝑜 𝑜𝑢𝑟 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
5𝑥
=
1
5
5−1
=
1
5
1
5
=
1
5
𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙, 𝑚𝑒𝑎𝑛𝑖𝑛𝑔 𝑜𝑢𝑟 𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡

9. Solving Exponential Equation
3) 6𝑥+2 = 36
6𝑥+2
= 62
(36 = 62
)
𝑥 + 2 = 2
𝑒𝑞𝑢𝑎𝑡𝑒 𝑡ℎ𝑒𝑖𝑟 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑠, 𝑡ℎ𝑒𝑛 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝒙
𝑥 = 2 − 2
𝑥 = 0
𝑁𝑜𝑤 𝑖𝑡′𝑠 𝑦𝑜𝑢𝑟 𝑡𝑖𝑚𝑒 𝑡𝑜 𝑐ℎ𝑒𝑐𝑘 𝑖𝑓 𝑡ℎ𝑒 𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡

10. Solving Exponential Equation
4) 93𝑥+1 = 81𝑥+1
93𝑥+1
= 92(𝑥+1)
(81 = 92
)
3𝑥 + 1 = 2𝑥 + 2
𝑒𝑞𝑢𝑎𝑡𝑒 𝑡ℎ𝑒𝑖𝑟 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑠, 𝑡ℎ𝑒𝑛 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝒙
𝑥 = 1
𝑁𝑜𝑤 𝑖𝑡′𝑠 𝑦𝑜𝑢𝑟 𝑡𝑖𝑚𝑒 𝑡𝑜 𝑐ℎ𝑒𝑐𝑘 𝑖𝑓 𝑡ℎ𝑒 𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡
93𝑥+1 = 92𝑥+2
3𝑥 − 2𝑥 = 2 − 1

11. Solving Exponential Inequality

12. In solving exponential inequality there are 2 rules that we need to consider:
• 𝑖𝑓 𝑏 > 1, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑦 = 𝑏𝑥 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥. 𝑇ℎ𝑖𝑠 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑎𝑡
𝑏𝑥 < 𝑏𝑦 𝒊𝒇 𝒂𝒏𝒅 𝒐𝒏𝒍𝒚 𝒊𝒇 𝑥 < 𝑦
• 𝑖𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑏 𝑖𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 1, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑖𝑠 𝒓𝒆𝒕𝒂𝒊𝒏𝒆𝒅
• 𝑖𝑓 𝑏 > 1, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑦 = 𝑏𝑥 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥. 𝑇ℎ𝑖𝑠 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑎𝑡
𝑏𝑥
< 𝑏𝑦
𝒊𝒇 𝒂𝒏𝒅 𝒐𝒏𝒍𝒚 𝒊𝒇 𝑥 < 𝑦
• 𝑖𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑏 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 1, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑖𝑠 𝒓𝒆𝒗𝒆𝒓𝒔𝒆𝒅
𝑹𝒖𝒍𝒆 #𝟏
𝑹𝒖𝒍𝒆 #𝟐
• 𝑖𝑓 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑜𝑓 𝑎𝑛 𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑎𝑟𝑒 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒊𝒆𝒅 𝒐𝒓 𝒅𝒊𝒗𝒊𝒅𝒆𝒅 𝑏𝑦 𝑎 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒓𝒆𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓, 𝑡ℎ𝑒𝑛
𝑡ℎ𝑒 𝑠𝑒𝑛𝑠𝑒 𝑜𝑓 𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑖𝑠 𝒓𝒆𝒕𝒂𝒊𝒏𝒆𝒅
• 𝑖𝑓 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑜𝑓 𝑎𝑛 𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑎𝑟𝑒 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒊𝒆𝒅 𝒐𝒓 𝒅𝒊𝒗𝒊𝒅𝒆𝒅 𝑏𝑦 𝑎 𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆 𝒓𝒆𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓, 𝑡ℎ𝑒𝑛
𝑡ℎ𝑒 𝑠𝑒𝑛𝑠𝑒 𝑜𝑓 𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑖𝑠 𝒓𝒆𝒗𝒆𝒓𝒔𝒆𝒅
(𝑜𝑛 𝑡ℎ𝑒 𝒑𝒓𝒐𝒄𝒆𝒔𝒔 𝑜𝑓 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥)
(𝑤ℎ𝑒𝑛 𝑓𝑖𝑛𝑑𝑖𝑛𝑔 𝑡ℎ𝑒 𝒇𝒊𝒏𝒂𝒍 𝒂𝒏𝒔𝒘𝒆𝒓 𝑓𝑜𝑟 𝑥)

13. Solving Exponential Inequality
1) 4𝑥+2 > 64
4𝑥+2 > 43
(64 = 43
)
𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝒙
𝑥 + 2 > 3 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 4 > 1, 𝒘𝒆 𝒓𝒆𝒕𝒂𝒊𝒏 𝒕𝒉𝒆 𝒔𝒊𝒈𝒏 >
𝑥 > 3 − 2
𝑥 > 1
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑠𝑒𝑡: (1, +∞)
𝑡ℎ𝑖𝑠 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑎𝑡 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑥 𝑎𝑟𝑒 𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 1
𝑡ℎ𝑒 𝑝𝑎𝑟𝑒𝑛𝑡ℎ𝑒𝑠𝑖𝑠 𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑒𝑠 𝑡ℎ𝑎𝑡 𝟏 𝑖𝑠 𝑛𝑜𝑡 𝑖𝑛𝑐𝑙𝑢𝑑𝑒𝑑 𝑎𝑠 𝑎 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥

14. Solving Exponential Inequality
2) 7𝑥 ≥ 49𝑥−3
7𝑥
≥ 72(𝑥−3)
(49 = 72
)
7𝑥 ≥ 72𝑥−6 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝒙
𝑥 ≥ 2𝑥 − 6 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 7 > 1, 𝒘𝒆 𝒓𝒆𝒕𝒂𝒊𝒏 𝒕𝒉𝒆 𝒔𝒊𝒈𝒏 ≥
𝑥 ≥ 2𝑥 − 6
𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑡𝑤𝑜 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝒙
𝑥 ≥ 2𝑥 − 6
𝑥 − 2𝑥 ≥ −6
−𝑥
−1
≥
−6
−1
𝑠𝑖𝑛𝑐𝑒 𝑤𝑒 𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦 𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆
𝑡ℎ𝑒 𝑠𝑖𝑔𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑖𝑠 𝒓𝒆𝒗𝒆𝒓𝒔𝒆𝒅
𝑥 ≤ 6
6 ≥ 2𝑥 − 𝑥
6 ≥ 𝑥
6 ≥ 𝑥
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑠𝑒𝑡: (−∞, 6]
𝑡ℎ𝑒 𝑏𝑟𝑎𝑐𝑘𝑒𝑡 𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑒𝑠 𝑡ℎ𝑎𝑡 𝟔 𝑖𝑠 𝑖𝑛𝑐𝑙𝑢𝑑𝑒𝑑 𝑎𝑠 𝑎 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥

15. Solving Exponential Inequality
3)
1
5
𝑥+2
≤
1
125
2𝑥
1
5
𝑥+2
≤
1
5
3(2𝑥)
(
1
125
=
1
5
3
)
1
5
𝑥+2
≤
1
5
6𝑥
𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝒙
𝑥 + 2 ≥ 6𝑥 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
1
5
< 1, 𝒘𝒆 𝒓𝒆𝒗𝒆𝒓𝒔𝒆 𝒕𝒉𝒆 𝒔𝒊𝒈𝒏 ≤ 𝑡𝑜 ≥
𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑡𝑤𝑜 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝒙
𝑥 + 2 ≥ 6𝑥
𝑥 − 6𝑥 ≥ −2
−5𝑥
−5
≥
−2
−5
𝑥 ≤
2
5
𝑠𝑖𝑛𝑐𝑒 𝑤𝑒 𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦 𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆 𝑡ℎ𝑒 𝑠𝑖𝑔𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑖𝑠 𝒓𝒆𝒗𝒆𝒓𝒔𝒆𝒅
𝑥 + 2 ≥ 6𝑥
2 ≥ 6𝑥 − 𝑥
2
5
≥
5𝑥
5
2
5
≥ 𝑥
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑠𝑒𝑡: (−∞,
2
5
]
𝑡ℎ𝑒 𝑏𝑟𝑎𝑐𝑘𝑒𝑡 𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑒𝑠 𝑡ℎ𝑎𝑡
2
5
𝑖𝑠 𝑖𝑛𝑐𝑙𝑢𝑑𝑒𝑑 𝑎𝑠 𝑎 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥

16. S.W. #5
A. Solve for the equality:
1) 52𝑥+2 = 125𝑥
2) 8𝑥−1 = 2𝑥+2
3)
1
3
𝑥+2
= 27−𝑥
4) 3𝑥2
= 9−2𝑥−3
B. Solve for the inequality:
5) 73𝑥+4
< 492𝑥+1
6) 44𝑥+10
≥ 642𝑥−2
7)
1
6
2𝑥−10
≤
1
36
3𝑥+13
8) 16𝑥+1 ≥ 44𝑥+1