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CIE 9709 2024 June Paper 31 Q2

A Level / CIE / P3

CIE 9709 2024 June Paper 31 Paper · Question 2

题目

Problem

Solve the equation ln⁡(x−5)=7−ln⁡x\ln(x-5)=7-\ln x. Give your answer correct to 22 decimal places.

(4)
题目中文翻译

解方程 ln⁡(x−5)=7−ln⁡x\ln(x-5)=7-\ln x。答案保留 22 位小数。

(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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先合并对数再指数化,得到二次方程。原对数要求 x>5x>5,因此只保留一个根。

答题过程

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The domain is x>5x>5. Rearranging,

ln⁡(x−5)+ln⁡x= 7,ln⁡[x(x−5)]= 7.\begin{align*} \ln(x-5)+\ln x=&\,7,\\[4mm] \ln[x(x-5)]=&\,7. \end{align*}

Thus

x(x−5)= e7,x2−5x−e7= 0.\begin{align*} x(x-5)=&\,e^7,\\[4mm] x^2-5x-e^7=&\,0. \end{align*}

Solving,

x=5±25+4e72.x=\frac{5\pm\sqrt{25+4e^7}}2.

The negative root is outside the domain. Hence

x= 5+25+4e72= 35.709684…≈35.71\begin{align*} x=&\,\frac{5+\sqrt{25+4e^7}}2\\[4mm] =&\,35.709684\ldots\approx\boxed{35.71} \end{align*}

to 2 decimal places.