## College Physics (4th Edition)

An intensity level change of $+1.00~dB$ corresponds to an increase in intensity of 25.9%
We can write an expression for $\beta_2-\beta_1 = 1.00~dB$ to find the change in intensity: $\beta_2-\beta_1 = 10~log~\frac{I_2}{I_0}-10~log~\frac{I_1}{I_0}$ $1.00 = 10~log~\frac{I_2}{I_1}$ $0.100 = log~\frac{I_2}{I_1}$ $10^{0.100} = \frac{I_2}{I_1}$ $I_2 = 10^{0.100}~I_1$ $I_2 = 1.259~I_1$ An intensity level change of $+1.00~dB$ corresponds to an increase in intensity of 25.9%.