Question

90db sound is x times more intense than 40db sound. what is x

Answer 1

I(dB) = 10 log₁₀ (I/I₀) where I₀ is a constant.
With 10 dB increase in intensity, the sound becomes 10 times more intense. Hence, with a 50 dB increase, the increase in intensity would be 10×10×10×10×10 = 10⁵. Hence, x = 10⁵.

Answer 2

The basic law governing this question is:
"For every 10db increase in sound, there is a 10-fold increase in sound intensity".

Thus, when we increase the sound from 40db to 50db, intensity increases by 10 times, and when we increase it from 50db to 60db, intensity increases by another 10 times, and so on.

From 40db to 90db, there is a 50db increase in sound. Thus, the intensity will increase 10 x 10 x 10 x 10 x 10 = $10^{5}$ times.

The value of x is therefore $10^{5}$

Hence,  90db sound is $10^{5}$ times more intense than 40db sound.