The decibel level of a sound is given by the equation D=10log 10 (i/i 0), where "i" is the sound's intensity and "i 0" corresponds to the intensity of the weakest sound that can be detected by the human ear. Approximately how many times greater than "i 0" must the intensity "i" of a sound be to reach 140 decibels?

Question

The decibel level of a sound is given by the equation D=10log 10 (i/i 0), where "i" is the sound's intensity and "i 0" corresponds to the intensity of the weakest sound that can be detected by the human ear. Approximately how many times greater than "i 0" must the intensity "i" of a sound be to reach 140 decibels?

A. 14 times greater
B. 1.14612 times greater
C. 10^14 times greater
D. 22026 times greater
E. 2.63906 times greater

Answer 1

10log 10 (i/i0) = 140

log (i/i0) = 14
i/i0 = 10^14

Answer 2

Well, let's unravel this equation, shall we? The intensity "i" of a sound is being compared to the intensity "i 0" of the weakest sound detectable by the human ear. So, we can say that "i" is "x" times greater than "i 0" to reach 140 decibels.

Using the equation D=10log 10 (i/i 0), we can rearrange it to find "x":
140 = 10 log 10 (i/i 0)
14 = log 10 (i/i 0)
10^14 = i/i 0 (applying the 10 to the power of both sides)

Therefore, the intensity "i" must be 10^14 times greater than "i 0" to reach 140 decibels. So the answer is...drumroll, please...option C – 10^14 times greater!

Answer 3

To find out how many times greater the intensity "i" must be compared to "i 0" to reach 140 decibels, we can rearrange the given equation and solve for "i/ i 0":

D = 10log 10 (i/ i 0)

By isolating (i/ i 0), we can rewrite the equation as:

(i/ i 0) = 10^(D/10)

Substituting D = 140 into the equation:

(i/ i 0) = 10^(140/10) = 10^14

Therefore, the intensity "i" must be 10^14 times greater than "i 0" to reach 140 decibels.

The correct answer is:
C. 10^14 times greater

Answer 4

To find out how many times greater the intensity "i" of a sound must be to reach 140 decibels, we can rearrange the given equation to solve for "i" in terms of the decibel level "D":

D = 10log10(i/i0)

First, we need to isolate "i" on one side of the equation:

D/10 = log10(i/i0)

Next, we can rewrite the equation in exponential form using the property of logarithms:

10^(D/10) = i/i0

Now, to find the number of times greater the intensity "i" must be compared to "i0", we divide both sides of the equation by "i0":

(10^(D/10))/i0 = i/i0

Since we are looking for the number of times greater, we can simplify the right side of the equation:

i/i0 = 10^(D/10)

Now we can substitute the given decibel level of 140 into the equation:

i/i0 = 10^(140/10)

Simplifying,

i/i0 = 10^14

Therefore, the intensity "i" of the sound must be 10^14 times greater than "i0" to reach 140 decibels.

The answer is C. 10^14 times greater.