📚 Example 1.8.2: Solving Logarithmic Equations | 示例1.8.2：解对数方程
This worked example is designed to reinforce your understanding of logarithmic equations, a key skill in A-Level Mathematics. Solving such equations requires a clear grasp of the logarithmic laws and the relationship between logarithms and exponentials.
本例题旨在帮助你巩固对数方程的求解能力，这是A-Level数学中的一项核心技能。求解这类方程需要清晰掌握对数运算律以及对数与指数之间的内在联系。
2. The Problem Statement | 题目描述
Consider the following equation, which is typical of what you may encounter in an AQA examination. Solve for x exactly:
请看下面这个方程，它属于你在AQA考试中可能遇到的典型题型。请精确求出 x 的值：
log₂(x+1) + log₂(x−1) = 3
At first glance, the equation involves two logarithmic terms with the same base. Our goal is to isolate the variable x by using logarithmic properties and then converting to exponential form.
乍一看，方程包含两个同底数的对数项。我们的目标是通过对数性质将变量 x 分离出来，然后转换为指数形式。
3. Key Logarithmic Rules | 对数关键法则
Before we begin solving, let us recall the essential rules that apply to logarithms with the same base. These rules are fundamental to simplifying the given expression.
在开始求解之前，我们先回顾一下同底数对数所适用的基本法则。这些法则对于简化给定表达式至关重要。
- Product rule: log_b(A) + log_b(B) = log_b(A×B)
- Exponential definition: log_b(A) = c ⇔ A = bᶜ
In this example, the product rule allows us to combine the two logarithms into a single logarithm. Then the definition of a logarithm lets us rewrite the equation without logarithms.
在本例中，乘积法则允许我们将两个对数合并为一个对数。然后，对数的定义使我们能够将方程改写为不含对数的形式。
4. Step 1: Combine the Logarithms | 第一步：合并对数
Apply the product rule to the left-hand side of the equation. Since both logarithms have base 2, we can write:
对方程左侧应用乘积法则。由于两个对数都以2为底，我们可以写作：
log₂[(x+1)(x−1)] = 3
Notice that the expression (x+1)(x−1) simplifies to x²−1 using the difference of two squares. This is a common algebraic simplification that appears frequently in logarithmic equations.
注意，利用平方差公式，(x+1)(x−1) 可以化简为 x²−1。这是对数方程中常见的一种代数化简技巧。
Therefore, the equation becomes:
因此，原方程变为：
log₂(x²−1) = 3
5. Step 2: Convert to Exponential Form | 第二步：转换为指数形式
Now that we have a single logarithm equal to a constant, we can use the definition of a logarithm. The statement log2(P) = 3 is equivalent to saying that P = 2³.
现在方程左侧只剩下一个对数等于常数，我们可以使用对数的定义。语句 log2(P) = 3 等价于 P = 2³。
Applying this to our equation gives:
将其应用到我们的方程中，得到：
x²−1 = 2³
Since 2³ = 8, we have:
由于 2³ = 8，于是：
x²−1 = 8
6. Step 3: Expand and Solve | 第三步：展开并解方程
Now solve the algebraic equation for x. Add 1 to both sides to isolate the squared term:
现在解这个代数方程，求出 x。方程两边同时加1，以分离出平方项：
x² = 9
Taking the square root of both sides, we must consider both the positive and negative roots:
对方程两边同时开平方，我们必须同时考虑正根和负根：
x = 3 or x = −3
However, we are not finished. The original logarithmic expression has domain restrictions. We must verify that each candidate satisfies the conditions under which the logarithms are defined.
然而，我们还没有完成。原始对数表达式存在定义域限制。我们必须验证每个候选解是否满足使对数有定义的条件。
7. Step 4: Check for Valid Solutions | 第四步：验证有效解
The original equation contains log2(x+1) and log2(x−1). For these to be defined, we require both arguments to be strictly positive:
原方程包含 log2(x+1) 和 log2(x−1)。为了使这些对数有定义，我们需要两个真数都严格为正：
x+1 > 0 and x−1 > 0
The second inequality implies x > 1, which automatically also makes x+1 > 0. Therefore the domain of the original equation is x > 1.
第二个不等式意味着 x > 1，这自然也使 x+1 > 0。因此原方程的定义域为 x > 1。
Now test the two candidate roots. Clearly, x = 3 satisfies x > 1, so it is a valid solution. On the other hand, x = −3 does not satisfy x > 1, and in fact both x+1 and x−1 would be negative, so log2(−2) and log2(−4) are undefined. Hence x = −3 must be rejected as an extraneous solution.
现在检验两个候选根。显然，x = 3 满足 x > 1，因此它是一个有效解。另一方面，x = −3 不满足 x > 1，实际上 x+1 和 x−1 都会变成负数，所以 log2(−2) 和 log2(−4) 无定义。因此必须将 x = −3 作为增根舍去。
Thus the only solution to the original equation is:
因此原方程的唯一解为：
x = 3
8. Common Pitfalls | 常见错误
Students often lose marks on logarithmic equations due to a few predictable mistakes. Being aware of them will help you avoid them in your own work.
在对数方程中，学生常常因为一些可预见的错误而丢分。了解这些错误将帮助你在自己的解答中避免它们。
- Forgetting the domain: Always check that your final answer makes every logarithm argument positive. Negative or zero arguments are invalid.
- Mixing up product and quotient rules: Remember that logb A + logb B = logb(AB), not logb(A+B).
- Applying the logarithm to each term separately: logb(A+B) cannot be split into logb A + logb B.
- Dropping the ± sign when taking square roots: Even though one root may be rejected, you must show both possibilities before discarding the invalid one.
In the next section, a similar exercise is provided for you to practice these skills independently.
在下一节中，我们提供了一个类似的练习，供你独立练习这些技能。
9. Practice Problem | 练习题
Solve the following equation for x, giving your answer in exact form where appropriate:
解下列方程，求出 x，答案请在合适时使用精确形式：
log₅(2x+3) − log₅(x−1) = 1
Hint: Use the quotient rule first, then convert to exponential form. Finally, check the domain of the original equation.
提示：先使用商法则，再转换为指数形式，最后检验原方程的定义域。
We encourage you to attempt this problem fully before checking your answer. The solution is x = 8/3, and it is valid because both arguments remain positive for that value.
我们鼓励你在查看答案之前完整尝试这道题。答案是 x = 8/3，该值有效，因为在这个值下两个真数都保持为正。
10. Summary | 总结
In this worked example, we solved the logarithmic equation log2(x+1) + log2(x−1) = 3 by following a clear four-step process: combine the logarithms using the product rule, convert to exponential form, solve the resulting algebraic equation, and finally check the validity of each root in the original domain.
在本例题中，我们按照清晰的四步流程求解了对数方程 log2(x+1) + log2(x−1) = 3：先用乘积法则合并对数，再转换为指数形式，然后解出所得代数方程，最后在原始定义域中验证每个根的有效性。
The key takeaway is that logarithmic equations require both algebraic manipulation and careful attention to domain restrictions. A solution that is algebraically correct may still be invalid if it lies outside the domain. Always finish with a verification step to ensure your final answer is mathematically sound.
关键的要点是：对数方程既需要代数变形，也需要仔细关注定义域限制。一个代数上正确的解，如果不在定义域内，仍然可能是无效的。务必以验证步骤作为收尾，以确保你的最终答案在数学上是严密的。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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