The plot of \( \log_{10} ({BMR}) \) as a function of \( \log_{10} (M) \) is a straight line with slope 0.75, where \( M \) is the mass of the person and BMR is the Basal Metabolic Rate. If a child with \( M = 10 \, {kg} \) has a BMR = 600 kcal/day, the BMR for an adult with \( M = 100 \, {kg} \) is _______ kcal/day. (rounded off to the nearest integer)
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When a logarithmic relationship is given, use the equation \( \log_{10} ({BMR}) = {slope} \cdot \log_{10} (M) + {intercept} \) to find the BMR for different masses.
We are given that \( \log_{10} ({BMR}) \) vs \( \log_{10} (M) \) is a straight line with a slope of 0.75. Therefore, we can write the equation of the line as: \[ \log_{10} ({BMR}) = 0.75 \cdot \log_{10} (M) + b \] Where: \( \log_{10} ({BMR}) \) is the logarithm of the Basal Metabolic Rate, \( \log_{10} (M) \) is the logarithm of the mass of the person, \( b \) is the y-intercept. We are also given that for a child with \( M = 10 \, {kg} \), \( {BMR} = 600 \, {kcal/day} \). Step 1: Determine the value of \( b \). Substitute \( M = 10 \) and \( {BMR} = 600 \) into the equation: \[ \log_{10} (600) = 0.75 \cdot \log_{10} (10) + b \] \[ \log_{10} (600) = 0.75 \cdot 1 + b \] \[ 2.778 = 0.75 + b \] \[ b = 2.028 \] Step 2: Find the BMR for \( M = 100 \, {kg} \). Now substitute \( M = 100 \) into the equation: \[ \log_{10} ({BMR}) = 0.75 \cdot \log_{10} (100) + 2.028 \] \[ \log_{10} ({BMR}) = 0.75 \cdot 2 + 2.028 \] \[ \log_{10} ({BMR}) = 1.5 + 2.028 = 3.528 \] Now, find the BMR: \[ {BMR} = 10^{3.528} \approx 3376.4 \, {kcal/day} \] Thus, the BMR for an adult with \( M = 100 \, {kg} \) is approximately **3376 kcal/day**.
