题目

Problem

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

(i) Using the laws of indices, solve the equation

\begin{align*} 4^{2y+1}=\frac{32^{4y}}{2} \end{align*}

(3)

(ii) Solve the equation

\begin{align*} x\sqrt{27}+21=\frac{6x}{\sqrt3} \end{align*}

writing the answer in the form $a\sqrt{b}$ where $a$ and $b$ are integers.

(4)

解答

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把 $4$ 和 $32$ 都写成以 $2$ 为底的幂。两边底数相同以后，就可以比较指数。

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\begin{align*} 4^{2y+1}=&\,\frac{32^{4y}}{2}\\ (2^2)^{2y+1}=&\,\frac{(2^5)^{4y}}{2}\\ 2^{4y+2}=&\,\frac{2^{20y}}{2}\\ 2^{4y+2}=&\,2^{20y-1} \end{align*}

Equating the indices gives

\begin{align*} 4y+2=&\,20y-1\\ 3=&\,16y\\ y=&\,\frac{3}{16}. \end{align*}

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先把 $\sqrt{27}$ 和 $\dfrac{6}{\sqrt3}$ 都化成 $\sqrt3$ 的倍数。这样整条方程就只剩同类根式，方便合并含 $x$ 的项。

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\begin{align*} \sqrt{27}=&\,3\sqrt3,\\ \frac{6}{\sqrt3} =&\,\frac{6\sqrt3}{3}\\ =&\,2\sqrt3. \end{align*}

\begin{align*} x\sqrt{27}+21=&\,\frac{6x}{\sqrt3}\\ 3\sqrt3x+21=&\,2\sqrt3x\\ \sqrt3x=&\,-21\\ x=&\,-\frac{21}{\sqrt3}\\ =&\,-\frac{21\sqrt3}{3}\\ =&\,-7\sqrt3. \end{align*}