Can someone help me? I'm really confused.

The loudness L of a sound in decibels is given by L= 10log(base10)R, where R is the sound's relative intensity. If the intensity of a certain sound is tripled, by how many decibels does the sound increase?

evaluate

L = 10log (3R)
= 10log3 + 10logR
= 4.77 + l0log R

so ....

oh, thanks

wait, where'd the 4.77 come from?

can u do the rest of the work??

My Name is Mac

Yes

To find out by how many decibels the sound increases when its intensity is tripled, let's consider the equation given:

L = 10 log10 R

Where L is the loudness in decibels and R is the relative intensity of the sound.

Now, we need to determine the increase in loudness when the intensity is tripled. Let's call the original intensity R1 and the tripled intensity R2.

From the given information, we know that R2 = 3R1 (the intensity is tripled).

By substituting the values into the equation, we get:

L2 = 10 log10 R2
L2 = 10 log10 (3R1)

Next, let's use a logarithmic property to simplify the expression:

L2 = 10 (log10 3 + log10 R1)

Since log10 3 is a constant value, let's denote it as K:

L2 = 10 (K + log10 R1)
L2 = 10K * 10(log10 R1)

Now, let's recall that log10 R1 was denoted as L in the initial equation (L = 10 log10 R).

So, we can write:
L2 = 10K * 10L

By combining the terms, we can express the increase in loudness in decibels as:

L2 - L = 10K * 10L - 10L

Simplifying this expression further:

L2 - L = (10K - 1) * 10L

Since K is a constant, 10K - 1 is also a constant. Let's denote it as M:

L2 - L = M * 10L

Therefore, we can conclude that when the intensity of a sound is tripled, the sound increases by M * 10L decibels.

Please note that the exact value of M can be calculated by substituting the constant value of log10 3 into the equation.