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IAL 2025 Jan Q7

A Level / Edexcel / P2

IAL 2025 Jan Paper · Question 7

题目

Problem

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

(i) The table below shows values of xx and yy, where y=log⁡10(x+5)y=\log_{10}(x+5), for xx values between −1-1 and 44

xx−1-10011223344
y=log⁡10(x+5)y=\log_{10}(x+5)log⁡104\log_{10}4log⁡105\log_{10}5log⁡106\log_{10}6log⁡107\log_{10}7log⁡108\log_{10}8log⁡109\log_{10}9

Using the trapezium rule with all the yy values in the given table, show that

∫−14log⁡10(x+5) dx≈log⁡10k\int_{-1}^{4}\log_{10}(x+5)\,dx\approx\log_{10}k

where kk is an integer to be found.

(3)

(ii) Find the value of aa such that

2log⁡5(5−a)−log⁡5(a+25)=12\log_5(5-a)-\log_5(a+25)=1
(5)

解答

(i)

解法一

思路

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这里步长是 11。用梯形法后，要用 log laws 把一串 log 合并成一个 log⁡10k\log_{10}k。

答题过程

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The step size is h=1h=1.

Using the trapezium rule,

∫−14log⁡10(x+5) dx≈12{log⁡104+log⁡109+2(log⁡105+log⁡106+log⁡107+log⁡108)}.\begin{align*} \int_{-1}^{4}\log_{10}(x+5)\,dx &\approx \frac12\{\log_{10}4+\log_{10}9\\ &\quad+2(\log_{10}5+\log_{10}6\\ &\quad+\log_{10}7+\log_{10}8)\}. \end{align*}

Now use log laws:

12{log⁡104+log⁡109+2(log⁡105+log⁡106+log⁡107+log⁡108)}= 12log⁡10(4⋅9)+log⁡10(5⋅6⋅7⋅8)= log⁡1036+log⁡10(1680)= log⁡106+log⁡101680= log⁡10(10080).\begin{align*} &\frac12\{\log_{10}4+\log_{10}9 +2(\log_{10}5+\log_{10}6\\ &\quad+\log_{10}7+\log_{10}8)\}\\ =&\,\frac12\log_{10}(4\cdot9) +\log_{10}(5\cdot6\cdot7\cdot8)\\ =&\,\log_{10}\sqrt{36} +\log_{10}(1680)\\ =&\,\log_{10}6+\log_{10}1680\\ =&\,\log_{10}(10080). \end{align*}

Therefore,

k=10080.\begin{align*} k=10080. \end{align*}

(ii)

解法一

思路

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先把 2log⁡5(5−a)2\log_5(5-a) 变成 log⁡5(5−a)2\log_5(5-a)^2，再把右边的 11 写成 log⁡55\log_5 5。最后注意 log 的定义域会排除 a=20a=20。

答题过程

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We have

2log⁡5(5−a)−log⁡5(a+25)= 1.\begin{align*} 2\log_5(5-a)-\log_5(a+25)=&\,1. \end{align*}

Using log laws,

log⁡5(5−a)2−log⁡5(a+25)= log⁡55log⁡5((5−a)2a+25)= log⁡55.\begin{align*} \log_5(5-a)^2-\log_5(a+25)=&\,\log_5 5\\[4mm] \log_5\left(\frac{(5-a)^2}{a+25}\right)=&\,\log_5 5. \end{align*}

Therefore,

(5−a)2a+25=  5(5−a)2=  5(a+25)a2−10a+25=  5a+125a2−15a−100=  0.\begin{align*} \frac{(5-a)^2}{a+25} =&\,\, 5\\[4mm] (5-a)^2 =&\,\, 5(a+25)\\[4mm] a^2-10a+25 =&\,\, 5a+125\\[4mm] a^2-15a-100 =&\,\, 0. \end{align*}

Factorise:

(a−20)(a+5)=  0.\begin{align*} (a-20)(a+5) =&\,\, 0. \end{align*}

So

a=20ora=−5.\begin{align*} a=20\quad\text{or}\quad a=-5. \end{align*}

But the original logarithms require

5−a>0\begin{align*} 5-a>0 \end{align*}

and

a+25>0.\begin{align*} a+25>0. \end{align*}

So a<5a<5 and a>−25a>-25. Hence a=20a=20 is not valid, and

a=−5.\begin{align*} a=-5. \end{align*}

解法二

思路

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对数定义移项合并法。将方程式左边的对数合并，写成单个对数式 log⁡5(5−a)2a+25=1\log_5 \frac{(5-a)^2}{a+25} = 1。然后直接根据对数定义将其转换为指数方程 (5−a)2a+25=51\frac{(5-a)^2}{a+25} = 5^1，避开将 11 写成对数的形式。

答题过程

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Start with the equation:

2log⁡5(5−a)−log⁡5(a+25)=1\begin{align*} 2\log_5(5-a)-\log_5(a+25)=1 \end{align*}

Apply the laws of logarithms:

log⁡5(5−a)2−log⁡5(a+25)=  1log⁡5((5−a)2a+25)=  1\begin{align*} \log_5(5-a)^2-\log_5(a+25) =&\,\, 1\\[4mm] \log_5\left(\frac{(5-a)^2}{a+25}\right) =&\,\, 1 \end{align*}

By the definition of a logarithm:

(5−a)2a+25=  51(5−a)2a+25=  5(5−a)2=  5(a+25)a2−10a+25=  5a+125a2−15a−100=  0.\begin{align*} \frac{(5-a)^2}{a+25} =&\,\, 5^1\\[4mm] \frac{(5-a)^2}{a+25} =&\,\, 5\\[4mm] (5-a)^2 =&\,\, 5(a+25)\\[4mm] a^2-10a+25 =&\,\, 5a+125\\[4mm] a^2-15a-100 =&\,\, 0. \end{align*}

Factorise:

(a−20)(a+5)=0  ⟹  a=20ora=−5\begin{align*} (a-20)(a+5) = 0 \implies a = 20 \quad\text{or}\quad a = -5 \end{align*}

Since the original logarithms require 5−a>0  ⟹  a<55-a > 0 \implies a < 5 and a+25>0  ⟹  a>−25a+25 > 0 \implies a > -25, we reject a=20a = 20.

Therefore, the only valid solution is:

a=−5.\begin{align*} a = -5. \end{align*}