Okay so the question is:
"Three people are cheering simultaneously. Measured individually, the cheers are 80dB, 80dB, and 50dB. What would the cheering measure totally? Use your knowledge of logarithms to solve."
So the formula we were given a few days ago was:
dB=10log(I/Io)
All I've done so far is plugged the dB values into the equation to get:
8=log(I1/Io), 8=log(I2/Io), and 5=log(I3/Io)
Where do I go from here? I'm lost :/
Great job on plugging in the values into the equation! Now, let's use the property of logarithms to simplify the equation further.
The property we will use is: log(A/B) = log(A) - log(B)
Applying this property to our equation, we get:
8 = log(I1/Io) becomes 8 = log(I1) - log(Io)
8 = log(I2/Io) becomes 8 = log(I2) - log(Io)
5 = log(I3/Io) becomes 5 = log(I3) - log(Io)
Now, let's rearrange these equations to isolate log(Io):
8 = log(I1) - log(Io) becomes log(Io) = log(I1) - 8
8 = log(I2) - log(Io) becomes log(Io) = log(I2) - 8
5 = log(I3) - log(Io) becomes log(Io) = log(I3) - 5
Now, we have three equations for log(Io). To solve, we can set the right-hand sides of these equations equal to each other:
log(I1) - 8 = log(I2) - 8 becomes log(I1) = log(I2)
log(I1) - 8 = log(I3) - 5 becomes log(I1) = log(I3) + 3
Since the logs on the left-hand sides of both equations are equal to log(I1), we can equate the right-hand sides:
log(I2) = log(I3) + 3
Now, we can apply the property of logarithms again to eliminate the logs:
I2 = I3 + 3
So, the intensity of the second person's cheer is 3 units higher than the intensity of the third person's cheer.
But we still need to find the total intensity when they cheer simultaneously. To do that, we need to add their intensities.
Using the formula: dB = 10log(I/Io), we can rearrange it to get I = 10^(dB/10):
For the first person's cheer, I1 = 10^(80/10) = 10^8
For the second person's cheer, I2 = 10^(80/10) = 10^8
For the third person's cheer, I3 = 10^(50/10) = 10^5
Now, to find the total intensity when they cheer simultaneously, we add their intensities:
Total intensity (Itotal) = I1 + I2 + I3
= (10^8) + (10^8) + (10^5)
= 200,000,000 + 200,000,000 + 100,000
= 400,000,000 + 100,000
= 400,100,000
Therefore, the total intensity of the cheering when they cheer simultaneously is 400,100,000.