Chapter 6.6, Problem 63E

a.

To determine

To rewrite the equation for $a$ in terms of $l$ .

Answer to Problem 63E

The equation in terms of $a$ is $a=-\frac{1}{0.09}\ln \left(\frac{45-l}{25.7}\right)$ .

Explanation of Solution

Given:

  $l=45-25.7e^{-0.09a}$

Calculation:

  $\begin{array}{c}l=45-25.7e^{-0.09a}\\ 25.7e^{-0.09a}=45-l\\ e^{-0.09a}=\frac{45-l}{25.7}\end{array}$

Take $\ln$ on both sides.

  $\begin{array}{c}\ln e^{-0.09a}=\ln \left(\frac{45-l}{25.7}\right)\\ -0.09a=\ln \left(\frac{45-l}{25.7}\right)\\ a=-\frac{1}{0.09}\ln \left(\frac{45-l}{25.7}\right)\end{array}$

b.

To determine

To find the ages of the elephants whose fingerprints are shown using part a.

Answer to Problem 63E

The ages are $2.2$ , $4.6$ , $7.6$ and $11.7$ years old respectively.

Explanation of Solution

Given:

The footprints of the elephants with length as $24\text{ cm}$ , $28\text{ cm}$ , $\text{32 cm}$ and $\text{36 cm}$ .

Calculation:

  $a=-\frac{1}{0.09}\ln \left(\frac{45-l}{25.7}\right)$

For $l=24\text{cm}$ ,

  $\begin{array}{c}a=-\frac{1}{0.09}\ln \left(\frac{45-24}{25.7}\right)\\ =-\frac{1}{0.09}\ln \left(0.817\right)\\ \approx 2.2\end{array}$

Thus, the age is $2.2\text{yrs}$ old.

For $l=28\text{cm}$ ,

  $\begin{array}{c}a=-\frac{1}{0.09}\ln \left(\frac{45-28}{25.7}\right)\\ =-\frac{1}{0.09}\ln \left(0.66\right)\\ \approx 4.6\end{array}$

Thus, the age is $4.6\text{yrs}$ old.

For $l=32\text{cm}$ ,

  $\begin{array}{c}a=-\frac{1}{0.09}\ln \left(\frac{45-32}{25.7}\right)\\ =-\frac{1}{0.09}\ln \left(0.505\right)\\ \approx 7.6\end{array}$

Thus, the age is $7.6\text{yrs}$ old.

For $l=36\text{cm}$ ,

  $\begin{array}{c}a=-\frac{1}{0.09}\ln \left(\frac{45-36}{25.7}\right)\\ =-\frac{1}{0.09}\ln \left(0.35\right)\\ \approx 11.7\end{array}$

Thus, the age is $11.7\text{yrs}$ old.