题目
Problem
Given that
3\log_2(t+4)-2\log_2(t-2)=7 \end{align*}$$ (a) verify that $t=4$ is a solution of the above equation, <div style="text-align: right;">(2)</div> (b) show that $$\begin{align*} t^3-116t^2+560t-448=0 \end{align*}$$ <div style="text-align: right;">(3)</div> (c) Hence, using algebra and showing your working, solve $$\begin{align*} 3\log_2(t+4)-2\log_2(t-2)=7 \end{align*}$$ giving each answer in simplest form. *(Solutions based entirely on calculator technology are not acceptable.)* <div style="text-align: right;">(4)</div>解答
(a)
解法一
思路
展开
把 代入方程两边，验证相等。
答题过程
展开
Substituting :
\text{LHS}=3\log_2(4+4)-2\log_2(4-2)=3\log_28-2\log_22=3\times3-2\times1=9-2=7=\text{RHS}.\quad\checkmark \end{align*}$$ </details> ## (b) ### 解法一 #### 思路 <details> <summary>展开</summary> 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。 </details> #### 答题过程 <details> <summary>展开</summary> Using log laws, $$\begin{align*} \log_2\!\left(\frac{(t+4)^3}{(t-2)^2}\right)=7 \end{align*}$$ $$\begin{align*} \frac{(t+4)^3}{(t-2)^2}=2^7=128 \end{align*}$$ $$\begin{align*} (t+4)^3=128(t-2)^2. \end{align*}$$ Expanding both sides: $$\begin{align*} t^3+12t^2+48t+64=128(t^2-4t+4) \end{align*}$$ $$\begin{align*} t^3+12t^2+48t+64=128t^2-512t+512 \end{align*}$$ $$\begin{align*} t^3-116t^2+560t-448=0.\quad\text{(shown)} \end{align*}$$ </details> ## (c) ### 解法一 #### 思路 <details> <summary>展开</summary> 由 (a) 知 $t=4$ 是根，用因式定理提出 $(t-4)$，再解二次方程。注意定义域 $t>2$。 </details> #### 答题过程 <details> <summary>展开</summary> From (a), $t=4$ is a root. Dividing $t^3-116t^2+560t-448$ by $(t-4)$: $$\begin{align*} t^3-116t^2+560t-448=(t-4)(t^2-112t+112)=0. \end{align*}$$ Setting each factor to zero: $$\begin{align*} t=4\qquad\text{or}\qquad t^2-112t+112=0. \end{align*}$$ For the quadratic, $$\begin{align*} t=\frac{112\pm\sqrt{12\,544-448}}{2}=\frac{112\pm\sqrt{12\,096}}{2}=\frac{112\pm24\sqrt{21}}{2}=56\pm12\sqrt{21}. \end{align*}$$ The domain requires $t+4>0$ and $t-2>0$, i.e.\ $t>2$. All three solutions satisfy this: $t=4>2$, $t=56-12\sqrt{21}\approx1.01$ (wait, let me check). $12\sqrt{21}\approx12\times4.583=54.99$, so $56-54.99\approx1.01<2$. This does **not** satisfy $t>2$, so it is rejected. $56+12\sqrt{21}\approx110.99>2$ ✓. $$\begin{align*} \boxed{t=4\quad\text{or}\quad t=56+12\sqrt{21}} \end{align*}$$ </details>