Chapter 6.6, Problem 17E

To determine

To calculate: .The age of a scalloped hammerhead shark.

Answer to Problem 17E

The required age of the shark is $17.56$ years.

Explanation of Solution

Given Information:

The given function is $l=266-219e^{-0.05t}$ .

Where $l$ is length and $t$ is the age (in years) of the shark.

Also $l$ is measure $175$ centimeters long.

Formula Used:

Power rule of logarithmic functions:

  ${\log }_a{x}^p=p{\log }_{a}x$

Where $a$ is the base, $x$ is the variable and $p$ is the power of variable.

Calculation:

Consider the function $l=266-219e^{-0.05t}$ .

Substitute $175$ for $l$ in the above equation.

  $175=266-219e^{-0.05t}$

Simplify the above equation.

  $\begin{array}{c}175=266-219e^{-0.05t}\\ 219e^{-0.05t}=266-175\\ 219e^{-0.05t}=91\end{array}$

Divide each side by 219 of the above equation.

  $\begin{array}{c}\frac{219e^{-0.05t}}{219}=\frac{91}{219}\\ e^{\left(-0.05t\right)}=0.416\end{array}$

Take natural log each side with base $e$ .

  ${\ln }_e{e}^{-0.05t}={\ln }_e\left(0.416\right)$

Now, use power rule of logarithm ${\log }_a{x}^p=p{\log }_{a}x$ .

  $\left(-0.05t\right){\ln }_{e}e={\ln }_e\left(0.416\right)$

Substitute $1$ for ${\ln }_{e}e$ in the above equations.

  $\begin{array}{c}\left(-0.05t\right)\left(1\right)={\ln }_e\left(0.416\right)\\ \left(-0.05\right)t={\ln }_e\left(0.416\right)\end{array}$

Substitute $-0.878$ for ${\ln }_e\left(0.416\right)$ in the above equation.

  $\left(-0.05\right)t=-0.878$

Simplify the above equation.

  $\begin{array}{c}\left(-0.05\right)t=-0.878\\ t=\frac{-0.878}{-0.05}\\ t=17.56\end{array}$

Hence, the required age of the shark is $17.56$ in years.