题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

(a) By substituting $p=3^x$ , show that the equation

\begin{align*} 3\times 9^x+3^{x+2}=1+3^{x-1} \end{align*}

can be rewritten in the form

\begin{align*} 9p^2+26p-3=0. \end{align*}

(3)

(b) Hence solve

\begin{align*} 3\times 9^x+3^{x+2}=1+3^{x-1}. \end{align*}

(3)

解答

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题目给了 $p=3^x$ ，所以要把所有指数项都改写成 $p$ 。关键是 $9^x=(3^2)^x=(3^x)^2=p^2$ ，而 $3^{x+2}=3^x\cdot3^2=9p$ 。

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Let

\begin{align*} p=3^x. \end{align*}

Then

\begin{align*} 9^x=&\,(3^2)^x=(3^x)^2=p^2,\\ 3^{x+2}=&\,3^x\cdot 3^2=9p,\\ 3^{x-1}=&\,\frac{3^x}{3}=\frac p3. \end{align*}

Substitute into the equation:

\begin{align*} 3p^2+9p=&\,1+\frac p3. \end{align*}

Multiply by $3$ :

\begin{align*} 9p^2+27p=&\,3+p\\ 9p^2+26p-3=&\,0. \end{align*}

This is the required form.

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先解上一小题得到的二次方程。因为 $p=3^x$ ，所以 $p$ 必须是正数，负根要舍去。

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From part (a),

\begin{align*} 9p^2+26p-3=&\,0. \end{align*}

Factorise:

\begin{align*} 9p^2+26p-3 =&\,(9p-1)(p+3). \end{align*}

\begin{align*} (9p-1)(p+3)=&\,0\\ p=&\,\frac19 \quad\text{or}\quad p=-3. \end{align*}

But $p=3^x>0$ , so $p=-3$ is not possible. Hence

\begin{align*} 3^x=&\,\frac19\\ 3^x=&\,3^{-2}. \end{align*}

Therefore

\begin{align*} x=-2. \end{align*}