a recent earthquake had a magnitude (r) of 3. what would the energy (e) of the earthquake be according to…

a recent earthquake had a magnitude (r) of 3. what would the energy (e) of the earthquake be according to the given model?\n$r = 0.67\\log(0.37e) + 1.46$\n$e = ?$ kwh\nround to the nearest hundredth.
Answer
Explanation:
Step1: Substitute R=3 into the model
$3 = 0.67\log(0.37E) + 1.46$
Step2: Isolate the logarithmic term
$0.67\log(0.37E) = 3 - 1.46$ $0.67\log(0.37E) = 1.54$ $\log(0.37E) = \frac{1.54}{0.67} \approx 2.2985$
Step3: Convert log to exponential form
$0.37E = 10^{2.2985} \approx 198.92$
Step4: Solve for E
$E = \frac{198.92}{0.37} \approx 537.62$
Answer:
$537.62$ kwh