# Really need help...

by Jayhawk1
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 HW Helper P: 2,566 The decibels of a magnitude is defined in relation to some base 0 dB level: $$V(dB) = 20 log(V/V_0)$$ But this will cancel out when finding the difference between the the decibel level of two magnitudes: $$V_2(dB) - V_1(dB) = 20 log(V_2/V_0) - 20 log(V_1/V_0) = 20 log((V_2/V_0)\cdot(V_0/V_1))=20 log(V_2/V_1)$$
 Sci Advisor HW Helper P: 3,033 You have the equation for this, I know. When you have something in a log that you have to figure out, you can eliminate the log by isolating it and using both side of the equation as a power of 10. The log is the inverse of 10 to a power a = b log(c) (a/b) = log(c) 10^(a/b) = c Use this on the db level equation you already have Specifically, $$L = 10log({\frac{I}{I_0}})$$ $$-15/10 = log({\frac{I}{I_0}})$$ $$10^{-1.5} = {\frac{I}{I_0}}$$ $$I_0 10^{-1.5} = I$$ You have the intensity at one point $$I_0$$. Now you can find it at the second point. Then you can calculate the distance you need to find. The factor 20 or 10 in the dB equation depends on whether you are talking about amplitude or energy. In your case, you are working with intensity (energy/area) so the factor is 10.
 Sci Advisor HW Helper P: 3,033 See the addition to the previous reply From above $$I_0 10^{-1.5} = I$$ $$.39 * 10^{-1.5} = I$$ $$.0123 = I$$ Now it is just like part a)