Comprehensive exercises on exponent calculations, graphs, equations, and inequalities.
Example ①
Find the minimum value of f(x)=9^x−2×3^x.
Answer: Let t=3^x（t>0）. f=t²−2t=(t−1)²−1. Minimum value −1（t=1、x=0）
Find the minimum value of f(x)=9^x−2×3^x.
Answer: Let t=3^x（t>0）. f=t²−2t=(t−1)²−1. Minimum value −1（t=1、x=0）
Example ②
When a^x=5 and a^y=3, find a^(2x−3y).
Answer: (a^x)²/(a^y)³ = 25/27
When a^x=5 and a^y=3, find a^(2x−3y).
Answer: (a^x)²/(a^y)³ = 25/27
Summary
- Extension of exponents: a^(m/n)=(ⁿ√a)ᵐ
- Graphs: increasing when a>1, decreasing when 0<a<1
- Equations and inequalities: set the bases equal→determine the direction of the inequality according to whether a>1
- Substitution: use t=2^x（t>0）to reduce the problem to a quadratic
Practice Problems
- When 2^x=3, find 8^x+8^(−x).
- Find the minimum value of f(x)=4^x−2^(x+1)+2.
- Solve 4^x−2^(x+3)+12≤0.
Answers and Explanations
- Answer: 730/27
Explanation: 8^x=(2^x)³=27、8^(−x)=1/27. Sum=27+1/27=730/27. - Answer: Minimum value 1（t=2^x、f=(t−1)²+1、the minimum occurs at t=1）
- Answer: 1≤x≤log₂6
Explanation: t=2^x（t>0）. 4^x−2^(x+3)+12=t²−8t+12≤0→(t−2)(t−6)≤0→2≤t≤6。Since t=2^x, 2≤2^x→x≥1、2^x≤6→x≤log₂6（≒2.58）. Therefore, 1≤x≤log₂6.