题目

Problem

The mass, $M$ kg, of a species of tree can be modelled by the equation

\log_{10}M=1.93\log_{10}r+0.684

where $r$ cm is the base radius of the tree.

The base radius of a particular tree of this species is $45$ cm.

According to the model,

(a) find the mass of this tree, giving your answer to 2 significant figures.

(2)

(b) Show that the equation of the model can be written in the form

M=pr^q

giving the values of the constants $p$ and $q$ to 3 significant figures.

(3)

(1)

题目中文翻译

某种树的质量 $M$ （单位：kg）可由方程

\log_{10}M=1.93\log_{10}r+0.684

建模，其中 $r$ cm 是树干底部半径。

某一棵这种树的底部半径是 $45$ cm。

根据该模型，

(a) 求这棵树的质量，答案保留 2 位有效数字。

(b) 证明模型方程可以写成

M=pr^q

的形式，并把常数 $p$ 与 $q$ 的值保留 3 位有效数字。

解答

Substitute $r=45$ into the model:

\log_{10}M=1.93\log_{10}45+0.684

\log_{10}M=3.8747\ldots

Therefore

M=10^{3.8747\ldots} =7492.9\ldots

To 2 significant figures,

\boxed{M=7500\text{ kg}}

Starting from

\log_{10}M=1.93\log_{10}r+0.684

we can write

\log_{10}M=\log_{10}(r^{1.93})+0.684

Also,

0.684=\log_{10}(10^{0.684})

\log_{10}M =\log_{10}(r^{1.93})+\log_{10}(10^{0.684})

Using the addition law of logarithms,

\log_{10}M =\log_{10}(10^{0.684}r^{1.93})

Hence

M=10^{0.684}r^{1.93}

Therefore

M=pr^q

where

p=10^{0.684}=4.831\ldots,\qquad q=1.93

So, to 3 significant figures,

\boxed{p=4.83,\qquad q=1.93}

In the model

M=pr^q

when $r=1$ ,

M=p

Therefore $p$ represents the predicted mass, in kg, of a tree of this species with base radius $1$ cm.