am I on the right track here?

A

allegansveritatem

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Here is the problem: (the problem referred to in #52 is mostly present in this image. The sentence that is not in the image is this: When the volume control on a stereo system is increased, the voltage across the loudspeaker....
dbproblem.PNG

I am trying to solve #52 for a+20. Here are my two tries:
dbtry1.PNG

and:
dbtry2.PNG

Am I on the path to resolution (or should I say salvation?) here?
 
1st line should not have db in it. What unit is db in (obvious)? What unit is 20 in? Should they be the same? So what should should db+20db equal? Answer 21db.

The lhs should just have 20, not db + 20.

Please don't just accept my advice but rather understand it.

For example let y equal the height. So in y = 2x+4, if we know that the height is 7 we do NOT write 7+y = 2x+4 but rather we write 7=x+4
 
Jomo said:
1st line should not have db in it. What unit is db in (obvious)? What unit is 20 in? Should they be the same? So what should should db+20db equal? Answer 21db.

The lhs should just have 20, not db + 20.

Please don't just accept my advice but rather understand it.

For example let y equal the height. So in y = 2x+4, if we know that the height is 7 we do NOT write 7+y = 2x+4 but rather we write 7=x+4
Click to expand...
yes, I see it. The plus sign threw me off...I thought they were saying: 20 plus what we already have. But now that I come to think of it, the plus sign is the convention when dealing with decibel readings above the zero mark. Thanks for pointing this out.
 
I disagree with the equation itself! "db" is the unit "decibels." It is not an Algebraic quantity! The equation should be
[math]I = 20 ~ log_{10} \left ( \dfrac{V_2}{V_1} \right )[/math]where I is the sound intensity measured in db. (Technically it should be [math]\Delta I[/math] as it's a change, but let's not quibble.)

Now you don't have to worry about the units... You are working with I + 20 db and the units take care of themselves.

-Dan
 
well, here is what I came up with today:
finally1.PNG

Seems to me it couldn't be anything else...but also it seems too neat and easy.
 
topsquark said:
I disagree with the equation itself! "db" is the unit "decibels." It is not an Algebraic quantity! The equation should be
[math]I = 20 ~ log_{10} \left ( \dfrac{V_2}{V_1} \right )[/math]where I is the sound intensity measured in db. (Technically it should be [math]\Delta I[/math] as it's a change, but let's not quibble.)

Now you don't have to worry about the units... You are working with I + 20 db and the units take care of themselves.

-Dan
Click to expand...
I will take your word for this. I just used 20 for the RS figure. But "I" is certainly what we are looking for.
 
topsquark said:
I disagree with the equation itself! "db" is the unit "decibels." It is not an Algebraic quantity! The equation should be
[math]I = 20 ~ log_{10} \left ( \dfrac{V_2}{V_1} \right )[/math]where I is the sound intensity measured in db. (Technically it should be [math]\Delta I[/math] as it's a change, but let's not quibble.)

Now you don't have to worry about the units... You are working with I + 20 db and the units take care of themselves.

-Dan
Click to expand...
I knew there was something that was bothering me. But with the OP's error I was just thrown off--way off. Thanks for pointing this out to me (and to the OP!)
 
allegansveritatem said:
well, here is what I came up with today:
View attachment 14803

Seems to me it couldn't be anything else...but also it seems too neat and easy.
Click to expand...
Why would you bring 20 to the power (although it is NOT wrong)? This is what bothers me with the way algebra is taught. You have 20*(something)=20 and most students are not taught to recognize that the something must be 1, even though they know as well as their own name that 20*1=20. They are taught to divide by 20! Thinking must be involved when teaching math!
Back to the problem logk= 1 and k=10 (since log10 =1) which is what you got. Good job!
 
Jomo said:
Why would you bring 20 to the power (although it is NOT wrong)? This is what bothers me with the way algebra is taught. You have 20*(something)=20 and most students are not taught to recognize that the something must be 1, even though they know as well as their own name that 20*1=20. They are taught to divide by 20! Thinking must be involved when teaching math!
Back to the problem logk= 1 and k=10 (since log10 =1) which is what you got. Good job!
Click to expand...
I'm not sure what you are saying here. I know that the 20 is a power--or as I think of it, an exponent--but ....I also see that there is a 20 on either side of the equation and both these twenties are powers. Now...well,one thing I am not clear on is this: I have some idea that one of the twentyies, even though it is in front of the log expression and multiplying it, still it is an integral part of the log expression and thus can't be arbitrarily divided from it for the purpose of dividing away something on the other side of the equation. I think that is why I did it the way I did.
 
[MATH]20 = 20log_{10}(K) \implies 1 = log_{10}(K) \implies K = 10^1 = 10.[/MATH]
Take the easy road: dividing by 20 isolates the log expression.
 
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