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

📚 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。

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1. Introduction | 引言

Exponential equations of the form a^2x + b · 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 · 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。

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2. The Problem | 题目陈述

Solve the equation 2^2x - 5 · 2^x + 6 = 0, where x is a real number.

求解方程 2^2x - 5 · 2^x + 6 = 0，其中 x 为实数。

2^2x − 5 × 2^x + 6 = 0

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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。

(2^x)² − 5 × 2^x + 6 = 0

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4. Step 2: Substitution | 第二步：换元

Let y = 2^x. Substituting into the equation gives y² - 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 y² - 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

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5. Step 3: Solving the Quadratic | 第三步：解二次方程

The quadratic y² - 5y + 6 factorises neatly as (y - 2)(y - 3) = 0. Hence y = 2 or y = 3.

二次方程 y² - 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

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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 = log₂ 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 = log₂ 3。

x = 1 或 x = log₂ 3

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7. Checking Solutions | 验证解

We verify each solution in the original equation. For x = 1, 2² - 5 · 2 + 6 = 4 - 10 + 6 = 0. For x = log₂ 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，2² - 5 × 2 + 6 = 4 - 10 + 6 = 0。当 x = log₂ 3 时，2^x = 3，所以 2^2x = 9；则 9 - 15 + 6 = 0。两个解均正确。

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8. Common Mistakes | 常见错误

• 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。

• Confusing 2^2x with 2^x2. They are different: 2^2x = (2^x)², not 2^x2.

  混淆 2^2x 与 2^x2。它们是不同的：2^2x = (2^x)²，而不是 2^x2。

• 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。

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9. Graphical Interpretation | 图形解释

The equation can be interpreted graphically by defining y = 2^2x - 5 · 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 × 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。

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10. Extension: Logarithmic Version | 扩展：对数形式

Suppose the equation were log₂² x - 5 log₂ x + 6 = 0. The same substitution works: let u = log₂ x, then u² - 5u + 6 = 0, giving u = 2 or u = 3. Hence x = 2² = 4 or x = 2³ = 8. This shows the power of the substitution method across different function types.

假设方程是 log₂² x - 5log₂ x + 6 = 0。同样的换元仍然有效：令 u = log₂ x，则 u² - 5u + 6 = 0，得 u = 2 或 u = 3。因此 x = 2² = 4 或 x = 2³ = 8。这说明换元法在不同函数类型中的强大作用。

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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：

• 3^2x - 10 · 3^x + 9 = 0

  3^2x - 10 × 3^x + 9 = 0

• 4^x - 3 · 2^x - 4 = 0

  4^x - 3 × 2^x - 4 = 0

• 2^2x - 6 · 2^x + 8 = 0

  2^2x - 6 × 2^x + 8 = 0

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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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