Math  /  Algebra

Question3. 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 logE=log10.61+0.1964logm\log E=\log 10.61+0.1964 \log m, where EE is the eye axial length, in millimetres, and mm is the body mass, in kilograms. To the nearest kilogram, predict the mass of a mountain goat with an eye axial length of 24 mm .

Studdy Solution

STEP 1

1. The relationship between eye size and body mass is given by the logarithmic equation: logE=log10.61+0.1964logm\log E = \log 10.61 + 0.1964 \log m.
2. We need to solve for m m when E=24 E = 24 mm.
3. The logarithm used is base 10.

STEP 2

1. Substitute the given eye axial length into the equation.
2. Solve the equation for logm\log m.
3. Calculate m m by exponentiating the result.
4. Round the result to the nearest kilogram.

STEP 3

Substitute E=24 E = 24 into the equation:
log24=log10.61+0.1964logm\log 24 = \log 10.61 + 0.1964 \log m

STEP 4

First, calculate log24\log 24 and log10.61\log 10.61:
log241.3802\log 24 \approx 1.3802 log10.611.0253\log 10.61 \approx 1.0253
Now substitute these values into the equation:
1.3802=1.0253+0.1964logm1.3802 = 1.0253 + 0.1964 \log m
Subtract log10.61\log 10.61 from both sides to isolate the term with logm\log m:
1.38021.0253=0.1964logm1.3802 - 1.0253 = 0.1964 \log m 0.3549=0.1964logm0.3549 = 0.1964 \log m

STEP 5

Solve for logm\log m by dividing both sides by 0.1964:
logm=0.35490.1964\log m = \frac{0.3549}{0.1964} logm1.806\log m \approx 1.806
Now, solve for m m by exponentiating both sides:
m=101.806m = 10^{1.806} m64.1m \approx 64.1

STEP 6

Round the calculated mass to the nearest kilogram:
m64 kgm \approx 64 \text{ kg}
The predicted mass of the mountain goat is:
64 kg \boxed{64} \text{ kg}

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