题目

Problem

Given that

\begin{align*} 3\log_2(t+4)-2\log_2(t-2)=&\,7 \end{align*}

(a) verify that $t=4$ is a solution of the above equation,

(2)

(b) show that

\begin{align*} t^3-116t^2+560t-448=&\,0 \end{align*}

(3)

\begin{align*} 3\log_2(t+4)-2\log_2(t-2)=&\,7 \end{align*}

giving each answer in simplest form.

(Solutions based entirely on calculator technology are not acceptable.)

(4)

题目中文翻译

已知

3\log_2(t+4)-2\log_2(t-2)=7

(a) 验证 $t=4$ 是上述方程的一个解。(2)

(b) 证明：(3)

t^3-116t^2+560t-448=0

3\log_2(t+4)-2\log_2(t-2)=7

（完全依赖计算器技术的解答不予接受。）

解答

(a)

解法一

思路

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把 $t=4$ 代入方程两边，验证相等。

答题过程

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Substituting $t=4$ :

\begin{align*} \text{LHS}=&\,3\log_2 8-2\log_2 2\\ =&\,3(3)-2(1)\\ =&\,7=\text{RHS}. \end{align*}

也可先合并对数再验证：

\begin{align*} &\,\log_2\frac{8^3}{2^2}\\ =&\,\log_2 128=7. \end{align*}

(b)

解法一

思路

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用对数法则合并： $\log_2\frac{(t+4)^3}{(t-2)^2}=7$ ，化为指数形式后展开整理。

答题过程

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Using log laws,

\begin{align*} \log_2\!\big(\frac{(t+4)^3}{(t-2)^2}\big)=&\,7 \end{align*}

\begin{align*} \frac{(t+4)^3}{(t-2)^2}=&\,2^7=128 \end{align*}

\begin{align*} (t+4)^3=&\,128(t-2)^2. \end{align*}

Expanding both sides:

\begin{align*} t^3+12t^2+48t+64=&\,128(t^2-4t+4) \end{align*}

\begin{align*} t^3+12t^2+48t+64\\ =&\,128t^2-512t+512 \end{align*}

\begin{align*} t^3-116t^2+560t-448=&\,0.\quad\text{(shown)} \end{align*}

(c)

解法一

思路

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承接 (a) 的根 $t=4$ 和 (b) 的三次方程，用多项式除法提出 $t-4$ 。解二次方程后必须回到原对数式的定义域筛选。

答题过程

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The original logarithms require $t+4>0$ and $t-2>0$ , so $t>2$ .

From (a), $t-4$ is a factor of the polynomial in (b). Let

F(t)=t^3-116t^2+560t-448.

Dividing by $t-4$ , the successive subtraction steps are

\begin{align*} F(t)-(t-4)t^2=&\,-112t^2+560t-448. \end{align*}

\begin{align*} &\,-112t^2+560t-448\\ &\,\hspace{2pt}-(t-4)(-112t)\\ =&\,112t-448. \end{align*}

112t-448-(t-4)(112)=0.

Thus

(t-4)(t^2-112t+112)=0.

Either $t=4$ , or

\begin{align*} t=&\,\frac{112\pm\sqrt{12544-448}}2\\ =&\,\frac{112\pm\sqrt{12096}}2\\ =&\,\frac{112\pm24\sqrt{21}}2\\ =&\,56\pm12\sqrt{21}. \end{align*}

Now $56-12\sqrt{21}\approx1.009<2$ , so it is rejected. The other two roots satisfy $t>2$ and hence the original equation. Therefore

t=4\quad\text{or}\quad t=56+12\sqrt{21}.

解法二

思路

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不用长除法，而是设商为首一二次式，通过比较系数求出它。之后配方求根，并按原式定义域筛选。

答题过程

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From (a) and (b), write

\begin{align*} &\,t^3-116t^2+560t-448\\ =&\,(t-4)(t^2+At+B). \end{align*}

Expanding the right-hand side gives

\begin{align*} &\,t^3+(A-4)t^2\\ &\,\hspace{2pt}+(B-4A)t-4B. \end{align*}

Comparing coefficients,

\begin{align*} A-4=&\,-116\ \Rightarrow\ A=-112\\ -4B=&\,-448\ \Rightarrow\ B=112. \end{align*}

The coefficient of $t$ also agrees:

B-4A=112+448=560.

Thus the equation is

(t-4)(t^2-112t+112)=0.

For the quadratic factor,

\begin{align*} t^2-112t+112=&\,0\\ (t-56)^2=&\,3024\\ t=&\,56\pm12\sqrt{21}. \end{align*}

Since the logarithms require $t>2$ , reject $56-12\sqrt{21}\approx1.009$ . Hence

t=4\quad\text{or}\quad t=56+12\sqrt{21}.

P2 2024 May R Q9

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程

解法二

思路

答题过程