## College Physics (4th Edition)

(a) Let's suppose that $I_2 = 10.0~I_1$. We can find the intensity level $\beta_2$: $\beta_2-\beta_1 = 10~log~\frac{I_2}{I_0}-10~log~\frac{I_1}{I_0}$ $\beta_2-\beta_1 = 10~log~\frac{I_2}{I_1}$ $\beta_2-\beta_1 = 10~log~\frac{10.0~I_1}{I_1}$ $\beta_2-\beta_1 = 10~log~10.0$ $\beta_2 - \beta_1 = 10.0$ $\beta_2 = \beta_1 + 10.0$ A factor of 10.0 increase in intensity corresponds to a 10.0-dB increase in intensity level. (2) Let's suppose that $I_2 = 2.0~I_1$. We can find the intensity level $\beta_2$: $\beta_2-\beta_1 = 10~log~\frac{I_2}{I_0}-10~log~\frac{I_1}{I_0}$ $\beta_2-\beta_1 = 10~log~\frac{I_2}{I_1}$ $\beta_2-\beta_1 = 10~log~\frac{2.0~I_1}{I_1}$ $\beta_2-\beta_1 = 10~log~2.0$ $\beta_2 - \beta_1 = 3.0$ $\beta_2 = \beta_1 + 3.0$ A factor of 2.0 increase in intensity corresponds to a 3.0-dB increase in intensity level.