Question

Small animal characters in animated features are often portrayed with big endearing eyes. In reality, the eye size of many vertebrates is related to body mass by the logarithmic equation \(\log E=\log 10.61+0.1964 \log m,\) where \(E\) is the eye axial length, in millimetres, and \(m\) is the body mass, in kilograms. To the nearest kilogram, predict the mass of a mountain goat with an eye axial length of \(24 \mathrm{mm}\)

Short answer

The body mass of a mountain goat with an eye axial length of 24 mm is approximately 64 kg.

Step-by-step solution

- Understand the Given Equation

The given logarithmic equation is \ \ \( \log E = \log 10.61 + 0.1964 \log m \) \ Where: \ \( E \) is the eye axial length, \( m \) is the body mass. Our task is to find the body mass \( m \) for an eye axial length \( E \) of 24 mm.

- Substitute the Given Value

Substitute \( E = 24 \) into the equation: \ \ \( \log 24 = \log 10.61 + 0.1964 \log m \)

- Evaluate \( \log 24 \) and \( \log 10.61 \)

Using a calculator, find \( \log 24 \) and \( \log 10.61 \): \ \ \( \log 24 \approx 1.3802 \) \ \( \log 10.61 \approx 1.0251 \)

- Substitute Logarithm Values

Substitute the calculated logarithm values back into the equation: \ \ \( 1.3802 = 1.0251 + 0.1964 \log m \)

- Isolate \( \log m \)

Subtract \( 1.0251 \) from both sides to isolate \( \log m \): \ \ \( 1.3802 - 1.0251 = 0.1964 \log m \) \ \ \( 0.3551 = 0.1964 \log m \)

- Solve for \( \log m \)

Divide both sides by 0.1964 to solve for \( \log m \): \ \ \( \log m = \frac{0.3551}{0.1964} \approx 1.808 \)

- Convert from Logarithmic Form

To find \( m \), convert from logarithmic form: \ \ \( m = 10^{1.808} \) \ Using a calculator, \( 10^{1.808} \approx 64.36 kg \)

- Round to the Nearest Kilogram

Round the body mass to the nearest kilogram: \ \ \( 64.36 \) rounds to \( 64 kg \)

Key concepts

logarithmic equations

Logarithmic equations are a powerful mathematical tool used to describe relationships where one quantity depends on the logarithm of another. In the context of biology, they often help to characterize how different biological parameters relate. The logarithmic equation given in the exercise is: \( \log E = \log 10.61 + 0.1964 \log m \). This equation suggests that the eye size \( E \) of vertebrates scales with their body mass \( m \) in a logarithmic manner. Here, \( E \) is the eye axial length in millimeters and \( m \) is the body mass in kilograms. Instead of a straightforward linear relationship, the logarithmic form accommodates more complex biological scaling laws.

vertebrate body mass

Understanding how body mass influences various biological attributes, such as eye size, can shed light on the principles of growth and development in vertebrates. Body mass is a fundamental biological parameter that affects an organism’s physiology. In this model, body mass \( m \) is a key predictor variable. As we have seen, the eye axial length \( E \) can be predicted using a specific logarithmic relationship with body mass. For instance, knowing the eye axial length of a particular vertebrate, like a mountain goat, allows us to estimate its body mass using the given logarithmic equation.

predicting from data

Predicting biological characteristics from numerical data relies on accurately substituting and solving equations. Here's a step-by-step approach: 1. Start with the known eye axial length. For the mountain goat, this length is \( 24 \) mm. 2. Replace \( E \) with \( 24 \) in the equation: \( \log 24 = \log 10.61 + 0.1964 \log m \). 3. Use a calculator to find \( \log 24 \approx 1.3802 \) and \( \log 10.61 \approx 1.0251 \). 4. Substitute these values back into the equation to isolate \( \log m \): \( 1.3802 = 1.0251 + 0.1964 \log m \). 5. Subtract \( 1.0251 \) from both sides: \( 1.3802 - 1.0251 = 0.1964 \log m \). 6. After isolating \( \log m \): \( 0.3551 = 0.1964 \log m \), divide by 0.1964: \( \log m = 1.808 \). 7. Convert \( \log m \) back to \( m \): \( m = 10^{1.808} \approx 64.36 \) kg. 8. Finally, round to the nearest kilogram: The mountain goat's body mass is approximately \( 64 \) kg.