📚 Solving log₂(x+1) + log₂(x-1) = 3, leading to x = 3 | 解对数方程 log₂(x+1) + log₂(x-1) = 3，求得 x = 3
ation log₂(x+1) + log₂(x-1) = 3, demonstrating how proper technique leads safely to x = 3.
对数方程在 A-Level Edexcel 数学中经常出现，需要熟练运用对数性质、定义域检查和代数变形。本文将逐步解方程 log₂(x+1) + log₂(x-1) = 3，展示严谨的方法如何自然地得出 x = 3。
1. Understanding Logarithmic Equations | 认识对数方程
m. To solve such equations, we rely on the definition of the logarithm: logₐ b = c means aˁ = b, where a > 0, a ≠ 1, and b > 0. This definition is the key to unlocking the unknown.
对数方程是指未知数出现在对数符号内部的方程。解这类方程需要依靠对数的定义：logₐ b = c 意味着 aˁ = b，其中 a > 0, a ≠ 1, 且 b > 0。这个定义是揭开未知数的关键。
n = logₐ (m n). This will help us reduce the equation to a single logarithmic expression.
当两个同底的对数相加时，如本题所示，我们通常可以利用乘积法则合并：logₐ m + logₐ n = logₐ (m n)。这有助于将方程简化为单一对数表达式。
2. The Given Equation | 给定方程
We are asked to solve the equation:
我们要求解如下方程：
log₂(x + 1) + log₂(x - 1) = 3
At first glance, it looks straightforward, but the steps must be taken with care because the arguments of the logarithms must remain positive. This imposes hidden restrictions on x.
乍一看这个方程很直接，但必须小心操作，因为对数的真数必须保持正数，这就对 x 施加了隐藏的限制条件。
3. Domain Restrictions | 定义域限制
Before manipulating the equation, always determine the domain. Since the arguments of the logs must be strictly positive, we require:
在对方程进行变形之前，必须先确定定义域。由于对数真数必须严格为正，我们需要：
- x + 1 > 0 → x > -1
- x - 1 > 0 → x > 1
Taking the intersection of these two conditions gives x > 1. This means any solution we find must satisfy x > 1, otherwise it is extraneous and must be discarded.
取这两个条件的交集，得到 x > 1。这意味着我们找到的任何解都必须满足 x > 1，否则就是增根，必须舍去。
4. Combining Logs Using the Product Rule | 使用乘积法则合并对数
With the domain noted, we can combine the two logarithms on the left-hand side using the product rule:
在明确了定义域之后，我们可以利用乘积法则将左边的两个对数合并：
log₂ ( (x + 1)(x - 1) ) = 3
This step is valid because both logs have the same base (2) and the arguments are positive under the domain x > 1.
这一步是有效的，因为两个对数具有相同的底数（2），并且在 x > 1 的前提下真数均为正。
5. Converting to Exponential Form | 转换成指数形式
Now we apply the definition of the logarithm. The equation log₂ (something) = 3 means:
现在我们应用对数的定义。方程 log₂ (某式) = 3 意味着：
(x + 1)(x - 1) = 2³
Since 2³ = 8, we obtain:
因为 2³ = 8，我们得到：
(x + 1)(x - 1) = 8
This step removes the logarithm and turns the equation into an algebraic one. Notice how the base 2 becomes the base of the power on the right.
这一步消除了对数符号，将方程转变为代数方程。注意底数 2 变成了右边幂的底数。
6. Expanding and Simplifying | 展开与化简
Next, we expand the left-hand side. Recognising that (x + 1)(x - 1) is a difference of squares, we can write:
接下来，我们展开左边。注意到 (x + 1)(x - 1) 符合平方差公式，可以写成：
x² - 1 = 8
Adding 1 to both sides gives:
两边加 1 得到：
x² = 9
At this point, we have a simple quadratic equation that leads to two potential solutions.
至此，我们得到一个简单的二次方程，它将产生两个可能的解。
7. Solving the Quadratic Equation | 解二次方程
Taking the square root of both sides yields:
两边开平方根得到：
x = ±√9 = ±3
-3. However, not both will satisfy the original logarithmic equation because of the domain restrictions we set earlier.
因此从数学上，我们得到 x = 3 和 x = -3。但是，并不是两个都能满足原对数方程，因为我们之前设定了定义域限制。
8. Checking Against Domain Restrictions | 对照定义域检查
Recall that the domain requires x > 1. Let us test each candidate:
回顾定义域要求 x > 1。我们来检验每个候选解：
- For x = 3: 3 > 1 → valid.
- For x = -3: -3 is not greater than 1 → invalid.
n extraneous solution. This illustrates why checking the domain is an essential part of solving logarithmic equations – skipping this step can lead to incorrect answers.
因此，x = -3 必须作为增根舍去。这说明为什么检查定义域是解对数方程的必要环节——跳过这一步可能导致错误答案。
9. Final Answer and Verification | 最终答案与验证
tion is x = 3. To be fully confident, we substitute x = 3 back into the original equation:
舍去增根后，我们得出结论：唯一有效解是 x = 3。为了更加确信，我们把 x = 3 代回原方程：
- 1) = log₂ 4 + log₂ 2 = 2 + 1 = 3 = RHS. The equality holds, so the solution is correct.
左边：log₂(3 + 1) + log₂(3 - 1) = log₂ 4 + log₂ 2 = 2 + 1 = 3 = 右边。等式成立，因此解是正确的。
Thus, the equation log₂(x+1) + log₂(x-1) = 3 leads to x = 3.
因此，方程 log₂(x+1) + log₂(x-1) = 3 的结果是 x = 3。
10. General Method and Key Points | 一般方法与要点总结
When tackling any logarithmic equation of this style, follow this systematic approach:
在解决此类对数方程时，请遵循以下系统方法：
- State the domain based on the arguments of the logarithms. Ensure each argument > 0.
- 根据真数列出定义域，保证每个真数 > 0。
- Combine log terms using log laws (product, quotient, or power rules) to obtain a single log on one side.
- 利用对数法则合并项（乘积、商、幂法则），使一边仅有一个对数。
- Rewrite in exponential form to eliminate the log.
- 重写为指数形式以消去对数。
- Solve the resulting algebraic equation (linear, quadratic, etc.).
- 解所得代数方程（线性、二次等）。
- Check each solution against the domain and reject any that fall outside. Verify by substitution if needed.
- 对照定义域检查每个解，舍去任何超出范围的解。必要时代入验证。
This method is robust and commonly examined in Edexcel A-Level Mathematics. Practise with varied bases and arguments to build fluency.
这种方法稳健可靠，在 Edexcel A-Level 数学中经常考查。多练习不同底数和真数的情况，以提高熟练度。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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