📚 Example 5.5.1: Solving Exponential Equations Using Substitution | 示例5.5.1：使用换元法求解指数方程
This article walks through a classic AQA A-Level Mathematics example that demonstrates how to solve an exponential equation by making a suitable substitution. The technique is essential for Paper 1 and Paper 2, and it frequently appears in non-calculator sections as well.
本文深入讲解一个经典的 AQA A-Level 数学例题，演示如何通过合理的换元来求解指数方程。该技巧在 Paper 1 和 Paper 2 中至关重要，也经常出现在不可使用计算器的部分。
1. Introduction | 引言
Exponential equations of the form (a^{2x} + b cdot a^x + c = 0) are not linear in (a^x), but they can be transformed into quadratic equations using a substitution. This approach simplifies the problem and makes the solution procedure systematic.
形如 (a^{2x} + b cdot a^x + c = 0) 的指数方程关于 (a^x) 并不是线性的，但通过换元可以将其转化为二次方程。这种方法简化了问题，使求解过程更加系统化。
2. The Problem | 题目陈述
Solve the equation (2^{2x} – 5 cdot 2^x + 6 = 0), where (x) is a real number.
求解方程 (2^{2x} – 5 cdot 2^x + 6 = 0)，其中 (x) 为实数。
22x − 5 × 2x + 6 = 0
3. Step 1: Rewriting the Equation | 第一步：重写方程
Notice that (2^{2x}) can be written as ((2^x)^2), because ((a^m)^n = a^{mn}). Therefore the original equation becomes a quadratic in the variable (y = 2^x).
注意 (2^{2x}) 可以写成 ((2^x)^2)，因为 ((a^m)^n = a^{mn})。因此原方程就变成了关于变量 (y = 2^x) 的二次方程。
(2x)2 − 5 × 2x + 6 = 0
4. Step 2: Substitution | 第二步：换元
Let (y = 2^x). Substituting into the equation gives (y^2 – 5y + 6 = 0). We must remember that (y > 0) because (2^x) is always positive for real (x).
令 (y = 2^x)。代入方程得 (y^2 – 5y + 6 = 0)。我们必须牢记 (y > 0)，因为对于实数 (x)，(2^x) 总是正的。
y2 − 5y + 6 = 0
5. Step 3: Solving the Quadratic | 第三步：解二次方程
The quadratic (y^2 – 5y + 6) factorises neatly as ((y – 2)(y – 3) = 0). Hence (y = 2) or (y = 3).
二次方程 (y^2 – 5y + 6) 可以顺利分解为 ((y – 2)(y – 3) = 0)。因此 (y = 2) 或 (y = 3)。
(y − 2)(y − 3) = 0 ⇒ y = 2 或 y = 3
6. Step 4: Back-Substitution | 第四步：代回原变量
Recall that (y = 2^x). For (y = 2), we have (2^x = 2), so (x = 1). For (y = 3), we have (2^x = 3), so (x = log_2 3).
回忆 (y = 2^x)。当 (y = 2) 时，有 (2^x = 2)，所以 (x = 1)。当 (y = 3) 时，有 (2^x = 3)，所以 (x = log_2 3)。
x = 1 或 x = log2 3
7. Checking Solutions | 验证解
We verify each solution in the original equation. For (x = 1), (2^{2} – 5 cdot 2 + 6 = 4 – 10 + 6 = 0). For (x = log_2 3), (2^x = 3), so (2^{2x} = 9); then (9 – 15 + 6 = 0). Both are correct.
我们将每个解代入原方程验证。当 (x = 1) 时，(2^{2} – 5 times 2 + 6 = 4 – 10 + 6 = 0)。当 (x = log_2 3) 时，(2^x = 3)，所以 (2^{2x} = 9)；则 (9 – 15 + 6 = 0)。两个解均正确。
8. Common Mistakes | 常见错误
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Forgetting that (2^x > 0) for all real (x), which would allow invalid negative solutions if they arose from the quadratic.
忘记 (2^x > 0) 对所有实数 (x) 成立，这会导致如果二次方程出现负根时接受无效的解。
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Confusing (2^{2x}) with (2^{x^2}). They are different: (2^{2x} = (2^x)^2), not (2^{x^2}).
混淆 (2^{2x}) 与 (2^{x^2})。它们是不同的：(2^{2x} = (2^x)^2)，而不是 (2^{x^2})。
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Dropping the factorisation step and trying to take logarithms directly, which makes the equation harder to solve.
跳过因式分解步骤而直接尝试取对数，这会使方程更难求解。
9. Graphical Interpretation | 图形解释
The equation can be interpreted graphically by defining (y = 2^{2x} – 5 cdot 2^x + 6). The two solutions correspond to the points where this curve crosses the x-axis. Since (2^x) grows rapidly, the graph has a U-shape in terms of (2^x), but is skewed when plotted against (x).
该方程可以通过定义 (y = 2^{2x} – 5 times 2^x + 6) 来作图解释。两个解对应于曲线与 x 轴的交点。由于 (2^x) 增长迅速，该曲线关于 (2^x) 呈 U 形，但在关于 (x) 的坐标平面中会倾斜。
10. Extension: Logarithmic Version | 扩展：对数形式
Suppose the equation were (log_2^2 x – 5 log_2 x + 6 = 0). The same substitution works: let (u = log_2 x), then (u^2 – 5u + 6 = 0), giving (u = 2) or (u = 3). Hence (x = 2^2 = 4) or (x = 2^3 = 8). This shows the power of the substitution method across different function types.
假设方程是 (log_2^2 x – 5log_2 x + 6 = 0)。同样的换元仍然有效：令 (u = log_2 x)，则 (u^2 – 5u + 6 = 0)，得 (u = 2) 或 (u = 3)。因此 (x = 2^2 = 4) 或 (x = 2^3 = 8)。这说明换元法在不同函数类型中的强大作用。
11. Practice Questions | 练习
Try solving these similar equations using the same technique:
请使用同样的技巧尝试求解以下类似方程：
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(3^{2x} – 10 cdot 3^x + 9 = 0)
(3^{2x} – 10 times 3^x + 9 = 0)
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(4^x – 3 cdot 2^x – 4 = 0)
(4^x – 3 times 2^x – 4 = 0)
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(2^{2x} – 6 cdot 2^x + 8 = 0)
(2^{2x} – 6 times 2^x + 8 = 0)
12. Summary | 总结
The key steps are: recognise the quadratic form, choose a positive substitution variable, solve the resulting quadratic, then convert back and check for validity. This method reliably solves many exponential equations tested in AQA A-Level Mathematics.
关键步骤为：识别二次形式，选择正的换元变量，求解所得二次方程，然后代回并检查有效性。这一方法可以可靠地解决 AQA A-Level 数学中许多指数方程问题。
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