1. ## Log problem

I have $\displaystyle L=10*log(x/I)$

Its a problem about sound levels where I = 10^-12 W/m^2 and L is counted in decibels (db)

The problem, what is x if L = 65 db

I know i should take the log of 65 somehow and i think * it by the log of I, but am not really sure

also in the picture the I has a lowewed 0 on it but i dont know how to right that or even what that means i have only seen raised ^

2. Originally Posted by JBswe
I have $\displaystyle L=10*log(x/I)$

Its a problem about sound levels where I = 10^-12 W/m^2 and L is counted in decibels (db)

The problem, what is x if L = 65 db

I know i should take the log of 65 somehow and i think * it by the log of I, but am not really sure

also in the picture the I is lowered to 0 but i dont know how to right that or even what that means i have only seen raised ^
$\displaystyle 65=10 \log_{10}(x) - 10 \log_{10}(I) = 10 \log_{10}(x) + 120$

so:

$\displaystyle -55=10 \log_{10}(x)$

or:

$\displaystyle x=10^{-5.5} \approx 3.16 \times~10^{-6} \ \rm{W/m^2}$

RonL

3. Hello, JBswe!

Give: .$\displaystyle L\:=\:10\cdot\log\left(\frac{x}{I_o}\right)$ . where: $\displaystyle I_o \:=\:10^{-12}$

If $\displaystyle L = 65$, find $\displaystyle x.$
The "sub-zero" usually means an initial value.
So that $\displaystyle I_o$ is some constant, not a variable.

We have: .$\displaystyle 10\cdot\log\left(\frac{x}{10^{-12}}\right) \;=\;65$

Divide by 10: .$\displaystyle \log\left(\frac{x}{10^{-12}}\right) \;=\;6.5$

Exponentiate both sides: .$\displaystyle \frac{x}{10^{-12}} \;=\;10^{6.5}$

Multiply by $\displaystyle 10^{-12}\!:\;\;x \;=\;10^{-12}\cdot10^{6.5}$

Therefore: .$\displaystyle x \;=\;10^{-5.5} \;\approx\;0.000003162$