MHB Comparing Pump to Dishwasher Intensity: 10x Difference

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The discussion focuses on comparing the sound intensity of a water pump and a dishwasher, with the pump at 50 decibels and the dishwasher at 62 decibels. Using the formula for sound intensity in decibels, the intensity ratio is calculated, revealing that the pump is approximately 0.063 times as intense as the dishwasher. This indicates that the dishwasher is significantly louder than the pump. The calculations involve converting decibel values into intensity ratios using logarithmic equations. The post gained attention on LinkedIn, attracting 14 views.
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$d=10\log\left({\frac{{P}_{}}{{P}_{0}}}\right)$
$d$=decibels
$P$ is intensity of the sound
$P_0$ is the weakest sound that the human ear can hear.

If a water pump is $d=50$ and a dishwasher $d=62$
How many times more intense is the pump to dishwasher.

Using exponential form then calculating

$\frac{{P}_{}}{{P}_{0}}=10^{50}$
and
$\frac{{P}_{}}{{P}_{0}}=10^{62}$
Now what?
 
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I would take:

$$d=10\log\left(\frac{P}{P_0}\right)$$

And solve it for $P$ to get:

$$P=P_010^{\frac{d}{10}}$$

And so we find the ratio $r$ of the intensity of the pump to that of the dishwasher to be:

$$r=\frac{P_010^{\frac{50}{10}}}{P_010^{\frac{62}{10}}}=10^{-1.2}\approx0.063095734448$$
 
i postered this on LinkedIn believing it might draw some into the forum it immediately got 14 views

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