[tex]\\ \rm\dashrightarrow B=10log\dfrac{I}{I_o}[/tex]
[tex]\\ \rm\dashrightarrow B=10\log\dfrac{10^{-7}}{10^{-12}}[/tex]
[tex]\\ \rm\dashrightarrow B=10log10^5[/tex]
[tex]\\ \rm\dashrightarrow B=10\times 5log10[/tex]
[tex]\\ \rm\dashrightarrow B=10(5)[/tex]
[tex]\\ \rm\dashrightarrow B=50dB[/tex]
Answer:
50 Db
Step-by-step explanation:
Given:
[tex]L=10 \log \dfrac{I}{I_0}[/tex]
[tex]I_0=10^{-12}[/tex]
where:
To find the approximate loudness of a dinner conversation with a sound intensity of [tex]I=10^{-7}[/tex], substitute the values of I and I₀ into the given equation and solve for L:
[tex]\implies L=10 \log \dfrac{10^{-7}}{10^{-12}}[/tex]
[tex]\textsf{Apply exponent rule} \quad \dfrac{a^b}{a^c}=a^{b-c}:[/tex]
[tex]\implies L=10 \log 10^{-7-(-12)[/tex]
[tex]\implies L=10 \log 10^5[/tex]
[tex]\textsf{Apply the log power law}: \quad \log_ax^n=n\log_ax[/tex]
[tex]\implies L=5 \cdot 10 \log 10[/tex]
[tex]\implies L=50 \log 10[/tex]
As the given log does not have a specified base, assume the base is 10.
[tex]\implies L=50 \log_{10} 10[/tex]
[tex]\textsf{Apply log law}: \quad \log_aa=1[/tex]
[tex]\implies L=50 (1)[/tex]
[tex]\implies L=50\:\:\sf Db[/tex]
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