Where on the x-axis are sound intensities equal between two sources?

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Two sources of sound are located on the x axis, and each unit emits power uniformly in all directions. There are no reflections. One source is positioned at the origin and the other at x = +164 m. The source at the origin emits four times as much power as the other source. (a) At which location between the two sources on the x-axis are the two sounds equal in intensity? (b) At which location to the right of the source at 164 m on the x-axis are the two sounds equal in intensity? Describe the locations by giving the distance from the origin.

I attempted to solve this multiple times and it is incorrect each and every time.

I made the origin I1, followed by the +164 I2, and I3 the location to the right

I set I1 and I2 equal to each other by

P1/4pi(164 + d)squared = P2/4pi(164 + d)squared

P2/P1= dsquared/ (164+d)squared

4P2/P1= dsquared/ (164+d)squared

1/4= (164-d)squared/ dsquared

(164+d/d)squared=(1/2)squared

164 + d/d= +-1/2

i solved and got d equal to -290 and -96.667 and these were wrongg so i don't know what i did
 

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xjessica said:
I made the origin I1, followed by the +164 I2, and I3 the location to the right

I'm not sure I understand what you mean by this. If there are only two sources, why do you need an [tex]I_3[/tex]? What is the intensity at the point x, due to (a)the source at the origin (b) the source at x=164 (c)both sources combined?

I set I1 and I2 equal to each other by

P1/4pi(164 + d)squared = P2/4pi(164 + d)squared

The method of setting [tex]I_1[/tex] and [tex]I_2[/tex] is correct, but your expressions for [tex]I_1[/tex] and [tex]I_2[/tex] are not. For [tex]I_1[/tex], is the distance from the origin to the point [tex]x=d[/tex] really [tex]164 +d[/tex]? Why wouldn't it just be [tex]{\Delta}x=d-0=d[/tex]? And for [tex]I_2[/tex] what is the distance from the point [tex]x=164[/tex] to the point [tex]x=d[/tex]?

P2/P1= dsquared/ (164+d)squared

4P2/P1= dsquared/ (164+d)squared

This doesn't look right, the proper way to simplify [tex]\frac{P_2}{P_1}[/tex] would be to substitute in the equation [tex]P_1=4P_2[/tex] as follows: [tex]\frac{P_2}{P_1}=\frac{P_2}{4P_2}=\frac{1}{4}[/tex]

Once you correct these mistakes, you should get the right answer. Try again and if you still have problems post your new attempt and I will go over it.