## College Physics (4th Edition)

To reduce the intensity by a factor of 2.0, the intensity level would have to be $97~dB$
Let $\beta_1 = 100.0~dB$ and suppose that $I_2 = \frac{I_1}{2.0}$. We can find the intensity level $\beta_2$: $\beta_1-\beta_2 = 10~log~\frac{I_1}{I_0}-10~log~\frac{I_2}{I_0}$ $\beta_1-\beta_2 = 10~log~\frac{I_1}{I_2}$ $\beta_1-\beta_2 = 10~log~\frac{I_1}{I_1/2.0}$ $\beta_1-\beta_2 = 10~log~2.0$ $\beta_2 = \beta_1 - 10~log~2.0$ $\beta_2 = 97~dB$ To reduce the intensity by a factor of 2.0, the intensity level would have to be $97~dB$.