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CIE 9709 2023 March Paper 32 Q1

A Level / CIE / P3

CIE 9709 2023 March Paper 32 Paper · Question 1

题目

Problem

It is given that x=ln⁡(2y−3)−ln⁡(y+4)x=\ln(2y-3)-\ln(y+4).

Express yy in terms of xx.

(3)
题目中文翻译

已知 x=ln⁡(2y−3)−ln⁡(y+4)x=\ln(2y-3)-\ln(y+4)。

用 xx 表示 yy。

(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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先确认原式中两个对数的真数都为正,再合并成商的对数,指数化并解出 yy。

答题过程

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The original logarithms require y>3/2y>3/2. Combining them,

x= ln⁡2y−3y+4,ex= 2y−3y+4.\begin{align*} x=&\,\ln\frac{2y-3}{y+4},\\[4mm] e^x=&\,\frac{2y-3}{y+4}. \end{align*}

Thus

ex(y+4)= 2y−3,(2−ex)y= 3+4ex.\begin{align*} e^x(y+4)=&\,2y-3,\\[4mm] (2-e^x)y=&\,3+4e^x. \end{align*}

Therefore

y=3+4ex2−ex.\boxed{y=\frac{3+4e^x}{2-e^x}}.

For y>3/2y>3/2, the original ratio lies between 00 and 22, so the corresponding range of xx is x<ln⁡2x<\ln2.