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CIE 9709 2025 June Paper 33 Q2

A Level / CIE / P3

CIE 9709 2025 June Paper 33 Paper · Question 2

题目

Problem

Solve the equation 2ln⁡(2x+3)−ln⁡(2x+5)=ln⁡(3x)2\ln(2x+3)-\ln(2x+5)=\ln(3x).

(4)
题目中文翻译

解方程 2ln⁡(2x+3)−ln⁡(2x+5)=ln⁡(3x)2\ln(2x+3)-\ln(2x+5)=\ln(3x)。

(4)
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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合并对数后得到二次方程,再回到定义域排除负根。

答题过程

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The logarithms require x>0x>0. Combining them,

ln⁡(2x+3)22x+5=ln⁡(3x).\ln\frac{(2x+3)^2}{2x+5}=\ln(3x).

Hence

(2x+3)2=3x(2x+5).(2x+3)^2=3x(2x+5).

Expanding and collecting terms,

4x2+12x+9= 6x2+15x,2x2+3x−9= 0,(2x−3)(x+3)= 0.\begin{align*} 4x^2+12x+9=&\,6x^2+15x,\\ 2x^2+3x-9=&\,0,\\ (2x-3)(x+3)=&\,0. \end{align*}

The root x=−3x=-3 is outside the domain. Therefore

x=32.\boxed{x=\frac32}.