# The volume control on a surround-sound amplifier is adjusted so the sound intensity level at the listening position increases from 23 to 61 dB. What is the ratio of the final sound intensity to the original sound intensity?

The volume control on a surround-sound amplifier is adjusted so the sound intensity level at the listening position increases from 23 to 61 dB.
What is the ratio of the final sound intensity to the original sound intensity?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

elilsonoulp2l
Given
${\beta }_{1}=23db$
${\beta }_{2}=61db$
The sound intensity level is given by
$\beta =10\mathrm{log}\left(\frac{I}{{I}_{0}}\right)$ where ${I}_{0}$ is intensity at reference level
Therefore the change in sound intensity level can be determined as
${\beta }_{2}-{\beta }_{1}=10\mathrm{log}\left(\frac{{I}_{2}}{{I}_{0}}\right)-10\mathrm{log}\left(\frac{{I}_{1}}{{I}_{0}}\right)$
$61-23=10\mathrm{log}\left(\frac{\frac{{I}_{2}}{{I}_{0}}}{\frac{{I}_{1}}{{I}_{0}}}\right)since\mathrm{log}a-\mathrm{log}b=\mathrm{log}\frac{a}{b}$
$38=10\mathrm{log}\left(\frac{{I}_{2}}{{I}_{1}}\right)$
$3.8=\mathrm{log}\left(\frac{{I}_{2}}{{I}_{1}}\right)$
$\frac{{I}_{2}}{{I}_{1}}={10}^{3.8}$
$\approx 6310$
Hence
The ratio of final sound intensity to original is approximately 6310
Result:
The ratio of final sound intensity to original is approximately 6310