Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.6 Applications and Models of Exponential Growth and Decay - 4.6 Exercises - Page 480: 22

$6.3$

Step 1. Given $M=6-\frac{5}{2}log(\frac{I}{I_0})$, we have $log(\frac{I}{I_0})=\frac{2}{5}(6-M)$, thus $I=I_0\cdot 10^{\frac{2}{5}(6-M)}$ Step 2. For $M=1$, we have $I_1=I_0\cdot 10^{\frac{2}{5}(6-1)}=100I_0$ Step 3. For $M=3$, we have $I_3=I_0\cdot 10^{\frac{2}{5}(6-3)}\approx15.85I_0$ Step 4. Thus $\frac{I_1}{I_3}\approx\frac{100I_0}{15.85I_0}\approx6.3$