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CIE 9709 2025 March Paper 32 Q1

A Level / CIE / P3

CIE 9709 2025 March Paper 32 Paper · Question 1

题目

Problem

Solve the equation

ln⁡(1−e−2x)+3=0.\ln(1-\mathrm{e}^{-2x})+3=0.

Give your final answer correct to 44 decimal places.

(3)
题目中文翻译

解方程

ln⁡(1−e−2x)+3=0.\ln(1-\mathrm{e}^{-2x})+3=0.

最终答案保留 44 位小数。

(3)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

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先移项并指数化,再解指数方程。原对数要求 x>0x>0,最后核对结果。

答题过程

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The logarithm requires 1−e−2x>01-e^{-2x}>0, so x>0x>0.

From the equation,

ln⁡(1−e−2x)=−3⟹1−e−2x=e−3.\ln(1-e^{-2x})=-3 \quad\Longrightarrow\quad 1-e^{-2x}=e^{-3}.

Thus

e−2x=1−e−3,e^{-2x}=1-e^{-3},

and taking logarithms gives

x=−12ln⁡(1−e−3)=0.0255\boxed{x=-\frac12\ln(1-e^{-3})=0.0255}

to 4 decimal places. This is positive and satisfies the original domain.