题目

Problem

Figure 1 shows a linear relationship between $\log_{10}y$ and $\log_{10}x$ .

The line passes through the points $(-3.5,0)$ and $(0,-2)$ as shown.

(a) Find an equation linking $\log_{10}y$ with $\log_{10}x$ .

(2)

(b) Hence, or otherwise, express $y$ in the form $px^q$ where $p$ and $q$ are rational constants.

(3)

题目中文翻译

图 1 显示了 $\log_{10}y$ 与 $\log_{10}x$ 之间的线性关系。

该直线经过点 $(-3.5,0)$ 和 $(0,-2)$ 。

(a) 求连接 $\log_{10}y$ 与 $\log_{10}x$ 的方程。

(b) 由此，或用其他方法，将 $y$ 表示成 $px^q$ 的形式，其中 $p,q$ 为有理常数。

解答

Let

Y=\log_{10}y,\qquad X=\log_{10}x

The line passes through

(-3.5,0)

and

(0,-2)

Its gradient is

m=\frac{-2-0}{0-(-3.5)} =\frac{-2}{3.5} =-\frac47

When

X=0

we have

Y=-2

So the equation is

Y=-\frac47X-2

Therefore

\boxed{\log_{10}y=-\frac47\log_{10}x-2}

From part (a),

\log_{10}y=-\frac47\log_{10}x-2

Use

-\frac47\log_{10}x=\log_{10}\left(x^{-\frac47}\right)

and

-2=\log_{10}(10^{-2})

Then

\log_{10}y =\log_{10}\left(x^{-\frac47}\right)+\log_{10}(10^{-2})

\log_{10}y =\log_{10}\left(10^{-2}x^{-\frac47}\right)

Therefore

y=10^{-2}x^{-\frac47}

Since

10^{-2}=\frac{1}{100}

we get

\boxed{y=\frac{1}{100}x^{-\frac47}}

Thus

\boxed{p=\frac{1}{100},\quad q=-\frac47}