Chapter 3.4, Problem 83E

To determine

Find the average diameter of the trees in a test plot for $N=21$ .

Answer to Problem 83E

The average diameter of the trees in a test plot for $N=21$ is $x=12.757$ .

Explanation of Solution

Given:

The given equation is $N=68\left({10}^{-0.04x}\right),5\le x\le 40$ . $x$ is the average diameter of the trees $3$ feet above the ground.

Calculation:

  $N=68\left({10}^{-0.04x}\right)$

Substitute the value of $N=21$ in the given equation.

  $\Rightarrow 21=68\left({10}^{-0.04x}\right)$

Divide both sides by $68$ .

  $\begin{array}{l}\Rightarrow \frac{21}{68}=\frac{68}{68}\left({10}^{-0.04x}\right)\\ \Rightarrow \frac{21}{68}={10}^{-0.04x}\\ \Rightarrow {10}^{-0.04x}=\frac{21}{68}\end{array}$

Taking log on both sides.

  $\Rightarrow \log {10}^{-0.04x}=\log \frac{21}{68}$

Use inverse property ${\log }_a{a}^x=x$

  $\begin{array}{l}\Rightarrow -0.04x=\log \frac{21}{68}\\ \Rightarrow -0.04x=\log \left(21\right)-\log \left(68\right)\\ \Rightarrow 0.04x=\log \left(68\right)-\log \left(21\right)\\ \Rightarrow 0.04x=1.83250-1.32221\\ \Rightarrow 0.04x=0.51029\end{array}$

Divide both sides by $0.04$ .

  $\begin{array}{l}\Rightarrow \frac{0.04}{0.04}x=\frac{0.51029}{0.04}\\ \Rightarrow x=\frac{0.51029}{0.04}\\ \Rightarrow x=12.757\end{array}$

Hence theaverage diameter of the trees in a test plot for $N=21$ is $x=12.757$ .