# Logarithmic and Exponential functions

• Aug 3rd 2012, 08:19 PM
astuart
Logarithmic and Exponential functions
Hello,

Not sure if this is in the right forum - feel free to move it.

I have a question from my textbook that is throwing me off a bit.

The relative loudness of a sound D of intensity I is measured in decibels (db) where

$D = 10 log I/Io$

and Io is the standard threshold of audibility.

A) Express the intensity of I of a 30-db sound (the sound level of normal conversation) in terms of Io.

I think I'm supposed to form an equation for I in relation to Io, but I'm not sure what I'm supposed to do. Apparently the answer is "10^3(Io)".

I'm thinking it's something to do with the fact that I = 30, and the fact that there's a 10 log expression - something to do with the 3 being a divisor of 30 (by 10). I'm just not sure how the 30 gets turned into the exponent of 10 (expressed as 3).

Any tips?
• Aug 3rd 2012, 08:35 PM
Prove It
Re: Logarithmic and Exponential functions
Quote:

Originally Posted by astuart
Hello,

Not sure if this is in the right forum - feel free to move it.

I have a question from my textbook that is throwing me off a bit.

The relative loudness of a sound D of intensity I is measured in decibels (db) where

$D = 10 log I/Io$

and Io is the standard threshold of audibility.

A) Express the intensity of I of a 30-db sound (the sound level of normal conversation) in terms of Io.

I think I'm supposed to form an equation for I in relation to Io, but I'm not sure what I'm supposed to do. Apparently the answer is "10^3(Io)".

I'm thinking it's something to do with the fact that I = 30, and the fact that there's a 10 log expression - something to do with the 3 being a divisor of 30 (by 10). I'm just not sure how the 30 gets turned into the exponent of 10 (expressed as 3).

Any tips?

Is this \displaystyle \begin{align*} \frac{10\log{I}}{I_0} \end{align*} or \displaystyle \begin{align*} 10\log{\left(\frac{I}{I_0}\right)} \end{align*}?
• Aug 3rd 2012, 08:37 PM
astuart
Re: Logarithmic and Exponential functions
Whoops, bit of a difference there. It's 10 log (I/Io)
• Aug 3rd 2012, 08:42 PM
Prove It
Re: Logarithmic and Exponential functions
Quote:

Originally Posted by astuart
Whoops, bit of a difference there. It's 10 log (I/Io)

Is the logarithm base 10 or base e?
• Aug 3rd 2012, 08:45 PM
astuart
Re: Logarithmic and Exponential functions
It doesn't specify, it's just written in the textbook as I've written above. I'm assuming it's log10, because previous questions have used 'ln' to denote log e.
• Aug 3rd 2012, 08:47 PM
Prove It
Re: Logarithmic and Exponential functions
Well then let D = 30 and transpose the equation so that you have \displaystyle \begin{align*} I_0 = \dots \end{align*}
• Aug 3rd 2012, 09:20 PM
Re: Logarithmic and Exponential functions
Quote:

Originally Posted by astuart
Hello,

Not sure if this is in the right forum - feel free to move it.

I have a question from my textbook that is throwing me off a bit.

The relative loudness of a sound D of intensity I is measured in decibels (db) where

$D = 10 log I/Io$

and Io is the standard threshold of audibility.

A) Express the intensity of I of a 30-db sound (the sound level of normal conversation) in terms of Io.

I think I'm supposed to form an equation for I in relation to Io, but I'm not sure what I'm supposed to do. Apparently the answer is "10^3(Io)".

I'm thinking it's something to do with the fact that I = 30, and the fact that there's a 10 log expression - something to do with the 3 being a divisor of 30 (by 10). I'm just not sure how the 30 gets turned into the exponent of 10 (expressed as 3).

Any tips?

While defining dB the base of logarithm is always 10.

$10 \log(\frac{I}{I_0})=30$

$\Rightarrow \log(\frac{I}{I_0})=3$

Hence by definition of log ie. if

[TEX]10^x=y[\TEX] Then [TEX]\log{(y)}=x [\TEX]

we get,

$\Rightarrow \frac{I}{I_0}=10^3$

$\Rightarrow I=10^3 I_0$
• Aug 3rd 2012, 09:45 PM
astuart
Re: Logarithmic and Exponential functions
Quote:

While defining dB the base of logarithm is always 10.

$10 \log(\frac{I}{I_0})=30$

$\Rightarrow \log(\frac{I}{I_0})=3$

Hence by definition of log ie. if

[TEX]10^x=y[\TEX] Then [TEX]\log{(y)}=x [\TEX]

we get,

$\Rightarrow \frac{I}{I_0}=10^3$

$\Rightarrow I=10^3 I_0$

Ah, that makes a bit more sense.

I was getting lost when I got to the point of $3 = log (I/Io)$. Didn't cross my mind to figure out the inverse.

If 3 = log (I/Io) then 1000 = (I/Io) = 10^3.

10^3 = (I/Io)
10^3(Io) = I

Cheers guys! :)