# You hear a sound at 65 dB. What is the

You hear a sound at 65 dB. What is the sound intensity level if the intensity of the sound is doubled?

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vrangett
Use the equation given by:
$$\displaystyle\beta={\left({10}{d}{B}\right)}{{\log}_{{{10}}}{\left({\frac{{{I}}}{{{I}_{{{0}}}}}}\right)}}$$
If I is doubled, use a property of logarithms to get the new value of the intensity level, $$\displaystyle\beta'$$. We have:
$$\displaystyle\beta'={\left({10}{d}{B}\right)}{{\log}_{{{10}}}{\left({\frac{{{2}{I}}}{{{I}_{{{0}}}}}}\right)}}{\left({10}{d}{B}\right)}{\left[{{\log}_{{{10}}}{\left({2}\right)}}+{{\log}_{{{10}}}{\left({\frac{{{I}}}{{{I}_{{{0}}}}}}\right)}}\right]}$$
$$\displaystyle={\left({10}{d}{B}\right)}{{\log}_{{{10}}}{2}}+\beta={3}{d}{B}+{65}{d}{B}={68}{d}{B}$$
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Durst37
Explanation:
What happens to sound intensity level when intensity levels are doubled is that one sounds twice as intense as the other. The sound of higher intensity has a sound level of about 3 dB higher.
The ratio of two intensities is 2 to 1, using of the properties of logarithms, $$\displaystyle{I}\frac{{2}}{{I}}{1}={2.00}$$
To show that the difference in sound levels is about 3 dB, that is,
$$\displaystyle\beta{2}-\beta{1}={3}{d}{B}$$.
Note that:
$$\displaystyle{\log{{10}}}{b}-{\log{{10}}}{a}={\log{{10}}}{\left({\frac{{{b}}}{{{a}}}}\right)}$$
$$\displaystyle\beta{2}-\beta{1}={10}{\log{{10}}}{\left({\frac{{{I}{2}}}{{{I}{1}}}}\right)}={10}{\log{{10}}}{\left({2.00}\right)}={10}{\left({0.301}\right)}{d}{B}$$
Thus, $$\displaystyle\beta{2}-\beta{1}={3.01}{d}{B}$$.
This means that the two sound intensity levels differ by 3.01 dB. Note that because only the ratio of the intensities is given (and not the actual intensities), this result is true for any intensities that differ by a factor of two. For instance a 68 dB sound is twice as intense as a 65 dB sound.