Jim halves the distance between himself and a sound source. What is the change in decibels of the sound he hears? -6 dB, -2 dB, +2 dB, or +6 dB ? :/

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Jim halves the distance between himself and a sound source. What is the change in decibels of the sound he hears? -6 dB, -2 dB, +2 dB, or +6 dB ? :/

Physics
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we start with an intensity I_1, after Jim halves the distance, the new intensity will be I_2= 4*I_1 so we can write: \[\large \begin{gathered} {N_1} = {\log _{10}}\left( {\frac{{{I_1}}}{{{I_0}}}} \right) \hfill \\ \hfill \\ {N_2} = {\log _{10}}\left( {\frac{{{I_2}}}{{{I_0}}}} \right) = {\log _{10}}\left( {\frac{{4 \times {I_1}}}{{{I_0}}}} \right) \hfill \\ \end{gathered} \] where I_0 is a reference intensity
ok! :)
what do we plug in from there?

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sorry the right formulas are: \[\large \begin{gathered} {N_1} = 10 \times {\log _{10}}\left( {\frac{{{I_1}}}{{{I_0}}}} \right) \hfill \\ \hfill \\ {N_2} = 10 \times {\log _{10}}\left( {\frac{{{I_2}}}{{{I_0}}}} \right) = 10 \times {\log _{10}}\left( {\frac{{4 \times {I_1}}}{{{I_0}}}} \right) \hfill \\ \end{gathered} \] now the requested change is: \[\large \begin{gathered} \Delta N = {N_2} - {N_1} = 10 \times {\log _{10}}\left( {\frac{{4 \times {I_1}}}{{{I_0}}}} \right) - 10 \times {\log _{10}}\left( {\frac{{{I_1}}}{{{I_0}}}} \right) = \hfill \\ \hfill \\ = 10 \times {\log _{10}}\left( 4 \right) \hfill \\ \end{gathered} \]
whoah! okay! so then we get 40?
no, since: \[\large 10 \times {\log _{10}}\left( 4 \right) = 10 \times 0.60206 = ...\]
ohh so it is 6.0206 so our answer is +6 dB?
yes! that's right!
yay!! thanks!!:)
:)

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