L(in dB)=10 logIIoL(in \ dB)=10 \ log\frac{I}{I_o}L(in dB)=10 logIoI
As per the question,I=Io2I=\frac{I_o}{2}I=2Io And substituting it in above equation,
L(in dB)=10 logIo2Io=10 log12L(in \ dB)=10 \ log\frac{I_o}{2I_o}=10 \ log\frac{1}{2}L(in dB)=10 log2IoIo=10 log21 =−3.010=-3.010=−3.010
This will be the change in the loudness level.
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