How many times more intense is $90\,\, dB$ sound than $40\,\, dB$ sound?
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Let $I$ represents the intensity of sound.

Loudness of sound $\quad L=10 \log _{10} \frac{I}{I_{o}}$

$\Longrightarrow L_{2}-L_{1}=10 \log _{10} \frac{I_{2}}{I_{1}}$

$90-40=10 \log _{10} \frac{I_{2}}{I_{1}}$

$5=\log _{10} \frac{I_{2}}{I_{1}} \Longrightarrow \frac{I_{2}}{I_{1}}=10^{5}$

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