Question

Help with Log models

Answer 1

The increase in sound intensity (density) is 48 - 45 = 3 dB.
Let the original intensity be P1 watts per square cm and the increased intensity be P2 watts per square cm.
We can write the following equation:
\[\large 3=10\log_{} \frac{P _{2}}{P _{1}}\ ..........(1)\]
Rearranging (1) gives:
\[\large \log_{} \frac{P _{2}}{P _{1}}=0.3\ ...........(2)\]
Do you follow so far?

Answer 2

.....increase in sound intensity (decibels)*

Answer 3

@kropot72 sorry I left last night

Answer 4

np. Shall we continue?

Answer 5

What did you do with the 10log in the first part is the onlything I dont understand

Answer 6

I divided both sides of equation (1) by 10 to get equation (2). Does that answer your query?

Answer 7

Oooo Ok thank you

Answer 8

Now we make use of the rules of logs as follows:
If log a = b, then
\[\large 10^{b}=a\]
Therefore
\[\large 10^{0.3}=\frac{P _{2}}{P _{1}}\ .........(3)\]
Does that make sense?

Answer 9

soh why is it exponent 0.3.

Answer 10

Look at equation (2).
0.3 is the log to base 10 of the intensity ratio.

Answer 11

So what is the value of 10^(0.3)?

Answer 12

1.99

Answer 13

Good, although 1.995 is required in this case, the reason being the answer must be rounded to the nearest tenth of a percent.
So the percentage intensity level is given by:
\[\large \frac{(1.995-1)}{1} \times\frac{100}{1}=you\ can\ calculate\]

Answer 14

the percentage intensity level increase is given by*

Answer 15

.995 X 100= 99.5?

Answer 16

99.5% is correct. A 3dB increase in power density corresponds to approximately doubling the power density.

Answer 17

Thank you

Answer 18

Your welcome :)

Answer 19

Could you help we with like 3 more doe .-. I really suck at dese

Answer 20

Sorry, I must log out now. Please post the others and someone else can help you.

Answer 21

Alright man thank you ar