Answer
The ratio of the greater intensity to the smaller intensity is $~~1.26$
Work Step by Step
Let $I_1$ be the smaller sound intensity.
We can find an expression for $I_1$:
$\beta_1 = (10~dB)~log~\frac{I_1}{I_0}$
$\frac{\beta_1}{10} = log~\frac{I_1}{I_0}$
$I_1 = 10^{\frac{\beta_1}{10}}~\cdot I_0$
Let $I_2$ be the greater sound intensity.
We can find an expression for $I_2$:
$\beta_1+1.00~dB = (10~dB)~log~\frac{I_2}{I_0}$
$\frac{\beta_1}{10}+0.10 = log~\frac{I_2}{I_0}$
$10^{(\frac{\beta_1}{10}+0.10)} = \frac{I_2}{I_0}$
$10^{0.10}~\cdot 10^{\frac{\beta_1}{10}} = \frac{I_2}{I_0}$
$I_2 = 10^{0.10}~\cdot10^{\frac{\beta_1}{10}}~\cdot I_0$
We can find the ratio of $\frac{I_2}{I_1}$:
$\frac{I_2}{I_1} = \frac{10^{0.10}~\cdot10^{\frac{\beta_1}{10}}~\cdot I_0}{10^{\frac{\beta_1}{10}}~\cdot I_0}$
$\frac{I_2}{I_1} = 10^{0.10}$
$\frac{I_2}{I_1} = 1.26$
The ratio of the greater intensity to the smaller intensity is $~~1.26$
