Answer:
[tex]\boxed{149.085}[/tex]
Step-by-step explanation:
Given the equation
[tex]D = 10 \log_{10} \left( \dfrac{I}{I_0}\right) \dots\dots[1][/tex]
where D = sound intensity in decibels
I = intensity of sound in watts per square meter
I₀ = 10⁻¹² watts per square meter
Given I = 8.1 x 10² watts per square meter we have to compute D
Plug in knowns into equation [1]
[tex]D = 10 \log_{10} \left( \dfrac{8.1 \cdot 10^2}{10^{-12}}\right)[/tex]
Let's first compute the term in the parenthesis:
- Apply exponent rule xᵃ/xᵇ = xᵃ⁻ᵇ
[tex]\dfrac{10^2}{10^{-12}} = 10^{2 - (-12)} = 10 ^ {2 + 12} = 10^{14}[/tex] - The term inside the parenthesis becomes:
[tex](8.1 \cdot 10^{14})[/tex] - Take log to base 10 of this value:
[tex]log_{10}\left(8.1 \times 10^{14}) = \log_{10}(8.1) + log_{10}({10^{14})[/tex]
This uses the log rule logₐ(xy) = logₐ(x) + logₐ(y) - Simplifying we get
[tex]log_{10}(8.1) = 0.908485\\log_{10}(10^{14}) = 14 \:\:\: \:\:\: \:\:\:\:\:\:\:\:\:[log_a(a^b) = b][/tex]
Expression becomes 0.908485 + 14 = 14.908485 - The D equation has 10 as a multiplier so
D = 10 (14.908485)
D = 149.08485
D = 149.085 rounded to 3 decimal places
