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P2 2024 Oct Q6

A Level / Edexcel / P2

IAL 2024 Oct Paper · Question 6

题目

Problem

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

Using the laws of logarithms, solve

log⁡4(12−2x)=2+2log⁡4(x+1)\log_4(12-2x)=2+2\log_4(x+1)
(5)
题目中文翻译

本题必须展示计算过程的所有步骤。

完全依赖计算器技术的解答不予接受。

使用对数法则，解以下方程。(5)

log⁡4(12−2x)=2+2log⁡4(x+1)\log_4(12-2x)=2+2\log_4(x+1)

解答

解法一

思路

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把 22 写成 log⁡416\log_4 16，把 2log⁡4(x+1)2\log_4(x+1) 写成 log⁡4((x+1)2)\log_4\big((x+1)^2\big)。最后要检查 log 定义域。

答题过程

展开 log⁡4(12−2x)= 2+2log⁡4(x+1)= log⁡416+log⁡4((x+1)2)= log⁡4{16(x+1)2}.\begin{align*} \log_4(12-2x) =&\, 2+2\log_4(x+1)\\ =&\, \log_4 16+\log_4\big((x+1)^2\big)\\ =&\, \log_4\{16(x+1)^2\}. \end{align*}

Therefore,

12−2x= 16(x+1)212−2x= 16x2+32x+1616x2+34x+4= 08x2+17x+2= 0.\begin{align*} 12-2x =&\, 16(x+1)^2\\ 12-2x =&\, 16x^2+32x+16\\ 16x^2+34x+4 =&\, 0\\ 8x^2+17x+2 =&\, 0. \end{align*}

Factorise:

(8x+1)(x+2)= 0.\begin{align*} (8x+1)(x+2) =&\, 0. \end{align*}

So

x= −18orx=−2.\begin{align*} x=&\,-\frac18\quad\text{or}\quad x=-2. \end{align*}

The logarithms require

12−2x>0\begin{align*} 12-2x>0 \end{align*}

and

x+1>0.\begin{align*} x+1>0. \end{align*}

Thus −1<x<6-1<x<6, so x=−2x=-2 is invalid. Therefore,

x= −18.\begin{align*} x=&\,-\frac18. \end{align*}

解法二

思路

展开

对数定义与移项合并法。将方程式右边的对数项移项到左边，利用对数的性质合并为单个对数式，然后应用对数定义（log⁡bN=c  ⟺  N=bc\log_b N = c \iff N = b^c）去掉对数符号，化为一元二次方程求解。

答题过程

展开

Rearrange the equation to group all logarithm terms on the left:

log⁡4(12−2x)−2log⁡4(x+1)= 2\begin{align*} \log_4(12-2x) - 2\log_4(x+1) =&\, 2 \end{align*}

Apply the laws of logarithms:

log⁡4(12−2x)−log⁡4((x+1)2)= 2log⁡4(12−2x(x+1)2)= 2\begin{align*} \log_4(12-2x) - \log_4\big((x+1)^2\big) =&\, 2\\ \log_4 \big( \frac{12-2x}{(x+1)^2} \big) =&\, 2 \end{align*}

By the definition of a logarithm:

12−2x(x+1)2= 4212−2x(x+1)2= 1612−2x= 16(x+1)212−2x= 16x2+32x+1616x2+34x+4= 08x2+17x+2= 0.\begin{align*} \frac{12-2x}{(x+1)^2} =&\, 4^2\\ \frac{12-2x}{(x+1)^2} =&\, 16\\ 12-2x =&\, 16(x+1)^2\\ 12-2x =&\, 16x^2+32x+16\\ 16x^2+34x+4 =&\, 0\\ 8x^2+17x+2 =&\, 0. \end{align*}

Factorise:

(8x+1)(x+2)= 0  ⟹  x=−18orx=−2\begin{align*} (8x+1)(x+2) =&\, 0 \implies x = -\frac18 \quad\text{or}\quad x = -2 \end{align*}

Since the domain of logarithms requires 12−2x>012-2x > 0 and x+1>0x+1 > 0, we have −1<x<6-1 < x < 6.

Thus, x=−2x = -2 is invalid.

Therefore, the only valid solution is:

x= −18.\begin{align*} x =&\, -\frac18. \end{align*}

解法三

思路

展开

先使用对数乘方法则，再对等式两边以 44 为底取指数，并用指数加法转乘法。解出的候选根仍须检查原对数定义域。

答题过程

展开

The original logarithms require −1<x<6-1<x<6.

Using the power law,

2log⁡4(x+1)=log⁡4((x+1)2).2\log_4(x+1)=\log_4\big((x+1)^2\big).

Raise 44 to both sides of the equation:

 4log⁡4(12−2x)= 42+log⁡4((x+1)2).\begin{align*} &\,4^{\log_4(12-2x)}\\ =&\,4^{2+\log_4((x+1)^2)}. \end{align*}

Using 4a+b=4a4b4^{a+b}=4^a4^b,

12−2x= 42⋅4log⁡4((x+1)2)= 16(x+1)2.\begin{align*} 12-2x=&\,4^2\cdot4^{\log_4((x+1)^2)}\\ =&\,16(x+1)^2. \end{align*}

Expanding and simplifying,

16x2+34x+4= 08x2+17x+2= 0(8x+1)(x+2)= 0.\begin{align*} 16x^2+34x+4=&\,0\\ 8x^2+17x+2=&\,0\\ (8x+1)(x+2)=&\,0. \end{align*}

The candidates are x=−18x=-\dfrac18 and x=−2x=-2. Reject −2-2 because x+1<0x+1<0. The only solution is

x=−18.x=-\frac18.