Chapter 3.6, Problem 59PS

Solution

To state: The comparison between the ages of two elephants, one with a shoulder height of 150 centimeters and the other with a shoulder height of 250 centimeters, biologists have discovered that the shoulder height h (in centimeters) of a male African elephant can be modeled by $h=62.5\sqrt[3]{t}+75.8$.

The elephant with a shoulder height of 250 cm is about 20 years older than the elephant with a shoulder height of 150 cm.

Given information:

The equation is $h=62.5\sqrt[3]{t}+75.8$ .

Formula used:

Raise each side of the equation to the same power.

Calculation:

Write the original equation:

  $h=62.5\sqrt[3]{t}+75.8$

Substitute the 150 into the equation,

  $150=62.5\sqrt[3]{t}+75.8$

Subtract 75.8 from both sides of the equation,

  $\begin{array}{c}150-75.8=62.5\sqrt[3]{t}+75.8-75.8\\ 74.2=62.5\sqrt[3]{t}\end{array}$

Now divide both sides by 62.5,

  $\begin{array}{c}\frac{74.2}{62.5}=\frac{62.5\sqrt[3]{t}}{62.5}\\ 1.19\approx \sqrt[3]{t}\end{array}$

Cube each side of the equation,

  $\begin{array}{c}{\left(1.19\right)}^3\approx {\left(\sqrt[3]{t}\right)}^3\\ t\approx 1.67\end{array}$

Therefore, the age of the elephant with a shoulder height of 150 centimeters is 1.67 years.

Now substitute the 250 into the equation,

  $250=62.5\sqrt[3]{t}+75.8$

Subtract 75.8 from both sides of the equation,

  $\begin{array}{c}250-75.8=62.5\sqrt[3]{t}+75.8-75.8\\ 174.2=62.5\sqrt[3]{t}\end{array}$

Now divide both sides by 62.5,

  $\begin{array}{c}\frac{174.2}{62.5}=\frac{62.5\sqrt[3]{t}}{62.5}\\ 2.79\approx \sqrt[3]{t}\end{array}$

Cube each side of the equation,

  $\begin{array}{c}{\left(2.79\right)}^3\approx {\left(\sqrt[3]{t}\right)}^3\\ t\approx 21.65\end{array}$

Therefore, the age of the elephant with a shoulder height of 250 centimeters is 21.65 years.