📚 Example 2.6.1: Solving Exponential Equations | 示例2.6.1：解指数方程
In this article, we work through a classic A-Level Mathematics example: solving an exponential equation where the unknown appears in the exponent of two different bases. The method we use is the natural logarithm, a fundamental technique required by the AQA specification.
本文将带领你完成一道经典的 A-Level 数学例题：解一个未知数同时出现在两个不同底数指数部分的指数方程。我们使用的方法是自然对数法，这也是 AQA 考纲中要求掌握的基本技能。
1. Problem Statement | 问题陈述
Consider the following equation, which corresponds to Example 2.6.1 in many textbooks:
以下方程即本题，对应许多教材中的 Example 2.6.1：
2x+1 = 3x-2
Our goal is to find the exact value of x and then approximate it to three significant figures.
我们的目标是求出 x 的精确值，并将其近似到三位有效数字。
2. Background Knowledge | 背景知识
Before solving, recall the key property of logarithms that makes this problem possible: for any positive number a ≠ 1 and any real number n,
在解题之前，请回顾对数的一个关键性质，正是它让这类问题得以求解：对于任意正数 a ≠ 1 和任意实数 n，
ln(aⁿ) = n ln a
This property allows us to bring the exponent down to the front, turning a non-linear exponential equation into a linear equation.
这个性质允许我们把指数放倒到前面，从而将一个非线性的指数方程转化为线性方程。
3. Taking Natural Logarithms | 取自然对数
Since both sides of the equation are positive for all real x, we may safely take the natural logarithm of both sides:
由于方程两边对于任意实数 x 均为正数，我们可以安全地对方程两边取自然对数：
ln(2x+1) = ln(3x-2)
Applying the power rule of logarithms to each side gives:
对两边应用对数的幂规则，得到：
(x + 1) ln 2 = (x - 2) ln 3
4. Expanding the Equation | 展开方程
Now we expand both sides by distributing the logarithmic factors:
现在，我们通过对数因子分别乘入括号中的各项，来展开方程的两边：
x ln 2 + ln 2 = x ln 3 - 2 ln 3
This step transforms the problem into a standard linear equation in x.
这一步将问题转化为关于 x 的标准线性方程。
5. Collecting Like Terms | 合并同类项
Move all terms containing x to one side and all constant terms to the other side:
将所有含 x 的项移到一边，所有常数项移到另一边：
x ln 2 - x ln 3 = -2 ln 3 - ln 2
Factor out x on the left-hand side:
在左边提出公因子 x：
x (ln 2 - ln 3) = -2 ln 3 - ln 2
6. Solving for x | 求解 x
Divide both sides by (ln 2 - ln 3). Note that this factor is not zero because ln 2 ≠ ln 3:
两边同时除以 (ln 2 - ln 3)。注意该因子不为零，因为 ln 2 ≠ ln 3：
x = (−2 ln 3 − ln 2) / (ln 2 − ln 3)
We can multiply the numerator and denominator by −1 to obtain a more elegant exact form:
我们可以将分子与分母同时乘以 −1，得到更简洁的精确形式：
x = (2 ln 3 + ln 2) / (ln 3 − ln 2)
This is the exact answer. Using the quotient rule for logarithms, it can also be written as:
这就是精确答案。利用对数的商规则，还可以写成：
x = ln 18 / ln(3/2)
because 2 ln 3 + ln 2 = ln 9 + ln 2 = ln 18.
因为 2 ln 3 + ln 2 = ln 9 + ln 2 = ln 18。
7. Numerical Approximation | 数值近似
Let us now compute a decimal value using a calculator. We know:
现在我们用计算器计算小数近似值。我们知道：
ln 3 ≈ 1.098612
ln 3 ≈ 1.098612
ln 2 ≈ 0.693147
ln 2 ≈ 0.693147
Therefore:
因此：
x ≈ (2 × 1.098612 + 0.693147) / (1.098612 − 0.693147) ≈ 2.890371 / 0.405465 ≈ 7.128
Rounded to three significant figures, we report x ≈ 7.13.
四舍五入到三位有效数字，我们得到 x ≈ 7.13。
8. Verification | 检验答案
We should always check our solution by substituting it back into the original equation.
我们应该始终坚持将解代回原方程进行检验。
For x = 7.128, the left-hand side is:
当 x = 7.128 时，左边为：
27.128+1 = 28.128 ≈ 279.5
The right-hand side is:
右边为：
37.128−2 = 35.128 ≈ 279.5
The two sides agree, so our solution is correct.
两边相等，所以我们的解是正确的。
9. Common Mistakes | 常见错误
Students often make several typical mistakes when solving such equations.
学生在解这类方程时经常犯几个典型错误。
Forgetting to take logs of both sides: Only applying the logarithm to one term breaks the equation. Always apply ln to both sides.
忘记对两边取对数：只对某一项取对数会破坏等式的平衡，一定要对方程两边同时取对数。
Incorrectly applying the power rule: Remember ln(2(x+1)) = (x+1) ln 2, not x ln 2 + ln 2 incorrectly? Actually that is correct because (x+1) ln 2 = x ln 2 + ln 2. The mistake is writing ln(2x+1) = ln 2x+1 without moving the exponent.
错误应用幂规则：牢记 ln(2(x+1)) = (x+1) ln 2，而不是其他错误的写法。常见错误是没有将指数拿到前面来。
Sign errors when rearranging: Moving terms across the equals sign requires changing their signs. Double-check each step.
移项时的符号错误：移项跨越等号时需要变号，每一步都要仔细检查。
Dividing by zero: The divisor ln 2 − ln 3 is nonzero, but students sometimes mistakenly divide by ln 2 − ln 3 without noticing it is safe. More generally, always check that you are not dividing by zero.
除以零：除数 ln 2 − ln 3 不为零，但学生有时会忽略这一点。更一般地，要时刻检查是否除以零。
10. Exam Tips | 考点总结
For AQA A-Level Mathematics, this type of question appears in both Paper 1 and Paper 2, often in the context of exponentials and logarithms.
在 AQA A-Level 数学考试中，这类题目出现在 Paper 1 和 Paper 2 中，通常结合指数与对数的知识点。
Always write down the exact answer before evaluating numerically. Marks are often awarded for the exact form.
在计算数值之前，先写出精确答案。分数往往授予精确形式。
Use the natural logarithm (ln) rather than log base 10, because ln is the standard in A-Level mathematics and simplifies differentiation later.
优先使用自然对数 ln，而不是常用对数 lg，因为 ln 是 A-Level 数学中的标准对数，且便于后续求导。
If the equation contains more than two exponential terms, try to isolate a single exponent term before taking logs.
如果方程包含两个以上的指数项，先尝试将其中一个指数项单独分离出来，再取对数。
n18 / ln(3/2) is acceptable, but you must also be able to produce a decimal approximation to the required accuracy.
请记住形如 x = ln18 / ln(3/2) 的答案是被接受的，但你也必须能给出符合精度要求的小数近似值。
11. Practice Problems | 变式练习
To master this technique, try solving the following similar equations by yourself.
为了掌握这一技巧，请自己尝试解下列类似的方程。
Solve 32x = 5x−1.
解方程 32x = 5x−1。
Solve 4x+2 = 71−x.
解方程 4x+2 = 71−x。
Solve 2x+3 · 3x = 62x−1.
解方程 2x+3 · 3x = 62x−1。
Each of these can be solved by taking natural logarithms and rearranging as shown above.
以上每道题都可以通过取自然对数并像上面一样重新整理来求解。
12. Conclusion | 小结
We have solved Example 2.6.1 by taking natural logarithms, expanding, collecting like terms, and solving for x. The key idea is to transform an exponential equation with different bases into a linear equation using logarithms.
我们通过取自然对数、展开、合并同类项并求解 x，完成了 Example 2.6.1。关键思想是利用对数，将不同底数的指数方程转化为线性方程。
Remember to present your final answer in both exact logarithmic form and a decimal approximation when requested. With careful practice, solving such equations will become quick and reliable.
请记住，在题目要求时，以精确的对数形式和十进制近似值两种方式呈现最终答案。通过仔细练习，解这类方程将变得快速且可靠。
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