Sound intensity increase in decibels

Text

The sound level in decibels is $L=10\,\log_{10}\dfrac{I}{I_0}$. If the intensity $I$ becomes $50$ times larger, by how many decibels does the sound level increase?

Solution

Let $L_1=10\log_{10}\dfrac{I}{I_0}$ and $L_2=10\log_{10}\dfrac{50\,I}{I_0}$. The increase is $\Delta L=L_2-L_1=10\left(\log_{10}\frac{50\,I}{I_0}-\log_{10}\frac{I}{I_0}\right)=10\,\log_{10}50.$ Since $\log_{10}50\approx 1.699$, $\Delta L\approx 10\cdot 1.699\approx 16.99\ \text{dB}.$ $\boxed{\,\Delta L\approx 16.99\ \text{dB}\,}$