# Thread: Log problem

1. ## Log problem

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

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 $L=10*log(x/I)$

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 ^
$
65=10 \log_{10}(x) - 10 \log_{10}(I) = 10 \log_{10}(x) + 120
$

so:

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

or:

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

RonL

3. Hello, JBswe!

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

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

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

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

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

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

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