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IAL 2025 May Q6

A Level / Edexcel / P2

IAL 2025 May Paper · Question 6

题目

Problem

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

(i) Solve

2log⁡2(4−x)=3+log⁡2(x+112)2\log_2(4-x)=3+\log_2\left(\frac{x+11}{2}\right)
(5)

(ii) The curves C1C_1 and C2C_2 with equations

y=32x+1andy=6×3xy=3^{2x+1} \qquad\text{and}\qquad y=6\times3^x

meet at the point PP.

Find the exact coordinates of PP, writing your answer in the form (log⁡3a,b)(\log_3 a,b) where aa and bb are integers.

(5)

解答

(i)

解法一

思路

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先用对数律把左边的 2log⁡2\log 合成平方，再把 33 写成 log⁡28\log_2 8。最后去掉对数，得到一个二次方程。别忘了检查对数定义域。

答题过程

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First use

2log⁡2(4−x)=log⁡2(4−x)2\begin{align*} 2\log_2(4-x)=\log_2(4-x)^2 \end{align*}

and

3=log⁡28.\begin{align*} 3=\log_2 8. \end{align*}

So

log⁡2(4−x)2= log⁡28+log⁡2(x+112)= log⁡2(8⋅x+112)= log⁡2(4x+44).\begin{align*} \log_2(4-x)^2 =&\,\log_2 8+\log_2\left(\frac{x+11}{2}\right)\\ =&\,\log_2\left(8\cdot\frac{x+11}{2}\right)\\ =&\,\log_2(4x+44). \end{align*}

Therefore,

(4−x)2=4x+44.\begin{align*} (4-x)^2=4x+44. \end{align*}

Solve:

16−8x+x2= 4x+44x2−12x−28= 0(x−14)(x+2)= 0.\begin{align*} 16-8x+x^2=&\,4x+44\\ x^2-12x-28=&\,0\\ (x-14)(x+2)=&\,0. \end{align*}

So

x=14orx=−2.\begin{align*} x=14\quad\text{or}\quad x=-2. \end{align*}

The logarithms require

4−x>0andx+112>0,4-x>0 \quad\text{and}\quad \frac{x+11}{2}>0,

so

−11<x<4.\begin{align*} -11<x<4. \end{align*}

Hence x=14x=14 is not allowed.

Therefore,

x=−2.\begin{align*} x=-2. \end{align*}

(ii)

解法一

思路

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交点处两个 yy 相等。把 32x+13^{2x+1} 写成 3⋅(3x)23\cdot(3^x)^2，再约去一个 3x3^x。

答题过程

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At the point of intersection,

32x+1=6×3x.\begin{align*} 3^{2x+1}=6\times3^x. \end{align*}

Now

32x+1=3(3x)2.\begin{align*} 3^{2x+1}=3(3^x)^2. \end{align*}

So

3(3x)2= 6(3x)3x= 2.\begin{align*} 3(3^x)^2=&\,6(3^x)\\ 3^x=&\,2. \end{align*}

Therefore,

x=log⁡32.\begin{align*} x=\log_3 2. \end{align*}

Substitute into y=6×3xy=6\times3^x:

y=6(2)=12.\begin{align*} y=6(2)=12. \end{align*}

Hence

P=(log⁡32,12).\begin{align*} P=(\log_3 2,12). \end{align*}

解法二

思路

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也可以直接对等式两边取以 33 为底的对数。这样能更清楚地把指数方程变成一次方程。

答题过程

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At the intersection,

32x+1=6×3x.\begin{align*} 3^{2x+1}=6\times3^x. \end{align*}

Taking log⁡3\log_3 of both sides gives

log⁡3(32x+1)= log⁡3(6×3x)2x+1= log⁡36+log⁡3(3x)2x+1= log⁡3(3⋅2)+x2x+1= 1+log⁡32+xx= log⁡32.\begin{align*} \log_3(3^{2x+1}) =&\,\log_3(6\times3^x)\\ 2x+1=&\,\log_3 6+\log_3(3^x)\\ 2x+1=&\,\log_3(3\cdot2)+x\\ 2x+1=&\,1+\log_3 2+x\\ x=&\,\log_3 2. \end{align*}

Then

y=6×3log⁡32=12.\begin{align*} y=6\times3^{\log_3 2}=12. \end{align*}

Therefore,

P=(log⁡32,12).\begin{align*} P=(\log_3 2,12). \end{align*}