Solution:
If the variation in pressure is P pounds per square inch, then the Loudness L in decibels is;
[tex]L=20\log _{10}(121.3P)[/tex]
When L=115 decibels;
[tex]\begin{gathered} 115=20\log _{10}(121.3P) \\ \text{Divide both sides by 20;} \\ \frac{115}{20}=\frac{20\log_{10}(121.3P)}{20} \\ \log _{10}(121.3P)=5.75 \end{gathered}[/tex]
But from the logarithmic law, we have;
[tex]\log _ba=c\leftrightarrow a=b^c[/tex]
Thus,
[tex]\begin{gathered} \log _{10}(121.3P)=5.75 \\ 121.3P=10^{5.75} \\ 121.3P=562341.33 \end{gathered}[/tex][tex]\begin{gathered} \text{Divide both sides by 121.3;} \\ \frac{121.3P}{121.3}=\frac{562341.33}{121.3} \\ P\cong4635.95 \end{gathered}[/tex]
FINAL ANSWER:
[tex]4636.0\text{ pounds per square inch.}[/tex]
