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IAL 2024 Jan Q6

A Level / Edexcel / P2

IAL 2024 Jan Paper · Question 6

题目

Problem

(a) Given that

2log4(x+3)+log4x=log4(4x+2)+122\log_4(x+3)+\log_4x=\log_4(4x+2)+\frac12

show that

x3+6x2+x4=0x^3+6x^2+x-4=0
(4)

(b) Given also that 1-1 is a root of the equation

x3+6x2+x4=0x^3+6x^2+x-4=0

(i) use algebra to find the other two roots of the equation.

(3)

(ii) Hence solve

2log4(x+3)+log4x=log4(4x+2)+122\log_4(x+3)+\log_4x=\log_4(4x+2)+\frac12
(1)

解答

(a)

解法一

思路

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先把 2log4(x+3)2\log_4(x+3) 写成 log4((x+3)2)\log_4((x+3)^2),再把 12\frac12 写成 log42\log_4 2。这样两边都可以合并成一个对数。

答题过程

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Start with

2log4(x+3)+log4x=log4(4x+2)+12.2\log_4(x+3)+\log_4x =\log_4(4x+2)+\frac12.

Using 2log4(x+3)=log4((x+3)2)2\log_4(x+3)=\log_4((x+3)^2) and 12=log42\frac12=\log_4 2,

log4((x+3)2)+log4x=log4(4x+2)+log42.\log_4((x+3)^2)+\log_4x =\log_4(4x+2)+\log_4 2.

So

log4(x(x+3)2)=log4(2(4x+2)).\log_4\left(x(x+3)^2\right) =\log_4(2(4x+2)).

Hence

x(x+3)2=2(4x+2)x(x2+6x+9)=8x+4x3+6x2+9x=8x+4x3+6x2+x4=0.\begin{align*} x(x+3)^2 =&\,\, 2(4x+2)\\[4mm] x(x^2+6x+9) =&\,\, 8x+4\\[4mm] x^3+6x^2+9x =&\,\, 8x+4\\[4mm] x^3+6x^2+x-4 =&\,\, 0. \end{align*}

This is the required result.

解法二

思路

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使用对数定义移项合并法。将所有含对数的项移项到左边,利用对数的性质合并为一个对数式,然后根据对数定义(logbN=c    N=bc\log_b N = c \iff N = b^c)去掉對數符號求解。

答题过程

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

2log4(x+3)+log4xlog4(4x+2)=12\begin{align*} 2\log_4(x+3)+\log_4x-\log_4(4x+2) = \frac12 \end{align*}

Apply the laws of logarithms:

log4(x+3)2+log4xlog4(4x+2)=12log4(x(x+3)24x+2)=12\begin{align*} \log_4(x+3)^2+\log_4x-\log_4(4x+2) =&\,\, \frac12\\[4mm] \log_4 \left( \frac{x(x+3)^2}{4x+2} \right) =&\,\, \frac12 \end{align*}

By the definition of a logarithm:

x(x+3)24x+2=41/2x(x+3)24x+2=2x(x+3)2=2(4x+2)x(x2+6x+9)=8x+4x3+6x2+9x=8x+4x3+6x2+x4=0.\begin{align*} \frac{x(x+3)^2}{4x+2} =&\,\, 4^{1/2}\\[4mm] \frac{x(x+3)^2}{4x+2} =&\,\, 2\\[4mm] x(x+3)^2 =&\,\, 2(4x+2)\\[4mm] x(x^2+6x+9) =&\,\, 8x+4\\[4mm] x^3+6x^2+9x =&\,\, 8x+4\\[4mm] x^3+6x^2+x-4 =&\,\, 0. \end{align*}

This is the required result.

(b)(i)

解法一

思路

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1-1 是根,所以 (x+1)(x+1) 是因式。把三次式除以 (x+1)(x+1),再解剩下的二次方程。

答题过程

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Since 1-1 is a root, (x+1)(x+1) is a factor.

Divide x3+6x2+x4x^3+6x^2+x-4 by (x+1)(x+1):

x3+6x2+x4=(x+1)(x2+5x4).\begin{align*} x^3+6x^2+x-4=(x+1)(x^2+5x-4). \end{align*}

Now solve

x2+5x4=0.\begin{align*} x^2+5x-4=0. \end{align*}

Using the quadratic formula,

x=5±524(1)(4)2=5±25+162=5±412.\begin{align*} x=&\,\frac{-5\pm\sqrt{5^2-4(1)(-4)}}{2}\\ =&\,\frac{-5\pm\sqrt{25+16}}{2}\\ =&\,\frac{-5\pm\sqrt{41}}{2}. \end{align*}

So the other two roots are

5+412and5412.\frac{-5+\sqrt{41}}{2} \quad\text{and}\quad \frac{-5-\sqrt{41}}{2}.

(b)(ii)

解法一

思路

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原来的对数方程要求 x>0x>0。三个根中只有 5+412\frac{-5+\sqrt{41}}2 是正数。

答题过程

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For the logarithmic equation, x>0x>0 is required.

The roots of the cubic are

1,5+412,5412.\begin{align*} -1,\quad \frac{-5+\sqrt{41}}{2},\quad \frac{-5-\sqrt{41}}{2}. \end{align*}

Only one of these is positive, so

x=5+412.\begin{align*} x=\frac{-5+\sqrt{41}}{2}. \end{align*}