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P2 2023 Oct Q5

A Level / Edexcel / P2

IAL 2023 Oct Paper · Question 5

题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

(i) Solve 3a=703^a = 70, giving the answer to 3 decimal places.

(2)

(ii) Find the exact value of bb such that

4+3log⁡3b=log⁡3(5b)4 + 3\log_3 b = \log_3(5b)
(4)
题目中文翻译

本题必须展示所有解题步骤，不能完全依赖计算器技术。

(i) 解 3a=703^a = 70，答案保留3位小数。

（2分）

(ii) 求满足 4+3log⁡3b=log⁡3(5b)4 + 3\log_3 b = \log_3(5b) 的 bb 的精确值。

（4分）

解答

(i)

解法一

思路

展开

两边取同底对数，再用幂法则和换底公式。

答题过程

展开

Solve 3a=703^a = 70:

log⁡3a= log⁡70alog⁡3= log⁡70a= log⁡70log⁡3≈3.867\begin{align*} \log 3^a =&\, \log 70 \\ a \log 3 =&\, \log 70 \\ a =&\, \frac{\log 70}{\log 3} \approx 3.867 \end{align*} a=3.867\boxed{a = 3.867}

(ii)

解法一

思路

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把对数差合并为一个商的对数，再转化为指数式。对数真数为正，故只取正根。

答题过程

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Solve 4+3log⁡3b=log⁡3(5b)4 + 3\log_3 b = \log_3(5b):

4+log⁡3b3= log⁡3(5b)4= log⁡3(5b)−log⁡3b34= log⁡35bb34= log⁡35b2\begin{align*} 4 + \log_3 b^3 =&\, \log_3(5b) \\ 4 =&\, \log_3(5b) - \log_3 b^3 \\ 4 =&\, \log_3 \frac{5b}{b^3} \\ 4 =&\, \log_3 \frac{5}{b^2} \end{align*}

Converting to index form:

34= 5b281= 5b2b2= 581b= 59\begin{align*} 3^4 =&\, \frac{5}{b^2} \\ 81 =&\, \frac{5}{b^2} \\ b^2 =&\, \frac{5}{81} \\ b =&\, \frac{\sqrt{5}}{9} \end{align*}

(Taking the positive value since the argument of a logarithm must be positive.)

b=59\boxed{b = \frac{\sqrt{5}}{9}}

解法二

思路

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先将常数 4 写成 log⁡381\log_3 81，把等号左侧合成一个乘积的对数，再比较真数。

答题过程

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Since the logarithms require b>0b>0,

log⁡381+log⁡3(b3)= log⁡3(5b)log⁡3(81b3)= log⁡3(5b).\begin{align*} \log_3 81+\log_3(b^3) =&\,\log_3(5b)\\ \log_3(81b^3) =&\,\log_3(5b). \end{align*}

Equating the arguments and dividing by b>0b>0:

81b3= 5b81b2= 5b2= 581.\begin{align*} 81b^3 =&\,5b\\ 81b^2 =&\,5\\ b^2 =&\,\frac5{81}. \end{align*}

Taking the positive root gives

b=59.\boxed{b=\frac{\sqrt5}{9}}.