D = 10 log ( ’ I ’ / 10⁻¹² )

D = 60. find ’ I ’.

Here we go:

60 = 10 log ( ’ I ’ / 10⁻¹² )

Divide each side by 10 :

6 = log ( ’ I ’ / 10⁻¹² )

Raise 10 to the power of each side of the equation:

10⁶ = ’ I ’ / 10⁻¹²

Multiply each side by 10¹² :

10¹⁸ = ’ I ’ That’s 10^18. It looks bad, because that isn’t one of the choices.

Let’s try a slightly different procedure:

============

After substituting 10⁺¹² for I₀, we’re working with this formula:

D = 10 log ( ’I’ / 10⁺¹² )

Let’s just look at the log part of that.

The log of a fraction is [ log(numerator) - log(denominator) ]

log of this fraction is [ log( ’I’ ) - log(10⁻¹²) ]

But log(10⁻¹²) is just (-12).

So the log of the fraction is [ log( ’I’ ) + 12 ]

And the whole formula is now:

D = 10 [ log( ’I’ ) + 12 ]

60 = 10 [ log( ’I’ ) + 12 ]

Divide each side by 10 :

6 = log( ’I’ ) + 12

Subtract 12 from each side :

-6 = log ( ’ I ’ )

’ I ’ = 10⁻⁶

That’s choice-’B’.

The equation for decibels and sound intensiv is \(D = 10\log_{10}\frac{i}{i_0} \), where D is decibels, I is intensity, and \(i_0=10^{-12}\) (the intensity of the finetest sound that can be heard by the ear). Find the sound intensity of normal conversation that mesures 60 decibels. A. \(10^{-60}\) B. \(10^{-6}\) C. \(10^{60}\) D.\(10^{6}\)

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