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![(10dB)log(\frac{I_{2} }{I_{o}} )-(10dB)log(\frac{I_{1} }{I_{o}} )=26.0dB\\(10dB)[log(\frac{I_{2} }{I_{o}} )-log(\frac{I_{1} }{I_{o}} )]=26.0dB\\log(\frac{\frac{I_{2}}{I_{o}} }{\frac{I_{1}}{I_{o}} })=\frac{26dB}{10dB}\\ log\frac{I_{2}}{I_{1}}=2.6\\ \frac{I_{2}}{I_{1}}=10^{2.6}](/tpl/images/0504/2021/bdb00.png)
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