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CIE 9709 2026 June Paper 33 Q1

A Level / CIE / P3

CIE 9709 2026 June Paper 33 Paper · Question 1

题目

Problem

Solve the equation ln⁡5+ln⁡(3x−2)=2ln⁡(x−4)\ln5+\ln(3x-2)=2\ln(x-4). Give your answer in a simplified exact form.

(4)
题目中文翻译

解方程 ln⁡5+ln⁡(3x−2)=2ln⁡(x−4)\ln5+\ln(3x-2)=2\ln(x-4)。答案写成化简后的精确形式。

(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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先检查对数的定义域,再合并对数。解二次方程后须排除不符合原定义域的根。

答题过程

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The logarithms require x>4x>4. Combining them gives

ln⁡(5(3x−2))=ln⁡((x−4)2).\ln\bigl(5(3x-2)\bigr)=\ln\bigl((x-4)^2\bigr).

Removing the logarithms and expanding gives

15x−10= x2−8x+16,x2−23x+26= 0.\begin{align*} 15x-10=&\,x^2-8x+16,\\[4mm] x^2-23x+26=&\,0. \end{align*}

The discriminant is 232−4(26)=425=25×1723^2-4(26)=425=25\times17, so the roots are (23±517)/2(23\pm5\sqrt{17})/2. The smaller is less than 44, so the solution is

x=23+5172.x=\frac{23+5\sqrt{17}}2.