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+2\log_4(x+1)

(5)

解答

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把 $2$ 写成 $\log_4 16$ ，把 $2\log_4(x+1)$ 写成 $\log_4(x+1)^2$ 。最后要检查 log 定义域。

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\begin{align*} \log_4(12-2x) =&\,\, 2+2\log_4(x+1)\\[4mm] =&\,\, \log_4 16+\log_4(x+1)^2\\[4mm] =&\,\, \log_4\{16(x+1)^2\}. \end{align*}

Therefore,

\begin{align*} 12-2x =&\,\, 16(x+1)^2\\[4mm] 12-2x =&\,\, 16x^2+32x+16\\[4mm] 16x^2+34x+4 =&\,\, 0\\[4mm] 8x^2+17x+2 =&\,\, 0. \end{align*}

Factorise:

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

\begin{align*} x=-\frac18\quad\text{or}\quad x=-2. \end{align*}

The logarithms require

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

and

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

Thus $x>-1$ , so $x=-2$ is invalid. Therefore,

\begin{align*} x=-\frac18. \end{align*}

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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（ $\log_b N = c \iff N = b^c$ ）去掉對數符號，化為一元二次方程求解。

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

\begin{align*} \log_4(12-2x) - 2\log_4(x+1) = 2 \end{align*}

Apply the laws of logarithms:

\begin{align*} \log_4(12-2x) - \log_4(x+1)^2 =&\,\, 2\\[4mm] \log_4 \left( \frac{12-2x}{(x+1)^2} \right) =&\,\, 2 \end{align*}

By the definition of a logarithm:

\begin{align*} \frac{12-2x}{(x+1)^2} =& \,\, 4^2\\[4mm] \frac{12-2x}{(x+1)^2} =& \,\, 16\\[4mm] 12-2x =&\,\, 16(x+1)^2\\[4mm] 12-2x =&\,\, 16x^2+32x+16\\[4mm] 16x^2+34x+4 =&\,\, 0\\[4mm] 8x^2+17x+2 =&\,\, 0. \end{align*}

Factorise:

\begin{align*} (8x+1)(x+2) = 0 \implies x = -\frac18 \quad\text{or}\quad x = -2 \end{align*}

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

Thus, $x = -2$ is invalid.

Therefore, the only valid solution is:

\begin{align*} x = -\frac18. \end{align*}