In a city,the sound intensity level increases | vedclass

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(C) The sound intensity level $L$ in decibels $(dB)$ is given by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$.Rearranging for intensit

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$3$

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(C) The sound intensity level $L$ in decibels $(dB)$ is given by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} ight)$.Rearranging for intensity $I$,we get: $I = I_0 \cdot 10^{L/10}$.Let $L_0$ be the initial sound level and $I_0$ be the initial intensity. After $t$ years,the sound level becomes $L = L_0 + t$ (since it increases by $1 \ dB$ per year).The new intensity $I'$ is given by: $I' = I_0 \cdot 10^{(L_0 + t)/10} = I_0 \cdot 10^{L_0/10} \cdot 10^{t/10}$.Since $I = I_0 \cdot 10^{L_0/10}$,we have $I' = I \cdot 10^{t/10}$.We want the intensity to double,so set $I' = 2I$:$2I = I \cdot 10^{t/10} \implies 2 = 10^{t/10}$.Taking the logarithm on both sides: $\log_{10}(2) = \frac{t}{10}$.Since $\log_{10}(2) \approx 0.3010$,we have $0.3010 = \frac{t}{10}$.Therefore,$t = 3.01 \approx 3 	ext{ years}$.

In a city,the sound intensity level increases uniformly by about $1 \ dB$ annually. How many years will it take for the intensity of sound to become double?

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