📚 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 \(\log_2^2 x – 5\log_2 x + 6 = 0\)。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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