When a person wears a hearing aid, the sound intensity level increases by 35 dB. By what factor does the sound intensity increase?
To determine the factor by which the sound intensity increases, we need to convert the increase in decibels (dB) to a factor.
The formula to calculate the increase in sound intensity in decibels is:
Increase in decibels = 10 * log10 (final intensity / initial intensity)
In this case, the increase in decibels is 35 dB, so we have:
35 = 10 * log10 (final intensity / initial intensity)
Now, let's solve for the factor by isolating the final intensity/initial intensity:
35/10 = log10 (final intensity / initial intensity)
3.5 = log10 (final intensity / initial intensity)
To remove the logarithm, we need to convert the equation to exponential form:
10^3.5 = final intensity / initial intensity
Using a calculator, 10^3.5 is approximately 3162.28.
Therefore, the factor by which the sound intensity increases when wearing a hearing aid is approximately 3162.28.
To determine the factor by which the sound intensity increases, we need to understand the relationship between sound intensity and decibels (dB).
The sound intensity level in decibels (L) is calculated using the equation:
L = 10 * log₁₀(I/I₀)
where I is the sound intensity being measured and I₀ is the reference sound intensity (typically set at the threshold of human hearing, which is 10⁻¹² W/m²).
In this case, we know that the sound intensity level increases by 35 dB. Let's call the initial sound intensity I₁ and the new sound intensity I₂.
So, we have the equation:
L₂ - L₁ = 35 dB
Substituting the equation for L into this equation:
10 * log₁₀(I₂/I₀) - 10 * log₁₀(I₁/I₀) = 35 dB
Now we can simplify the equation:
log₁₀(I₂/I₀) - log₁₀(I₁/I₀) = 3.5
Using the logarithmic property log a - log b = log (a/b), the equation becomes:
log₁₀((I₂/I₀) / (I₁/I₀)) = 3.5
Simplifying further:
log₁₀(I₂/I₁) = 3.5
Using the logarithmic property log a = b if and only if a = 10^b, we can convert the equation into exponential form:
I₂/I₁ = 10^(3.5)
Now we can solve for the factor by taking the antilog of both sides:
I₂/I₁ = 3162.3
Therefore, the sound intensity increases by a factor of 3162.3 when a person wears a hearing aid.