Intensity of sound wave question The question is:
The intensity of sound wave A is 100 times weaker than that of sound wave B. Relative to wave B the sound level of wave A is?
The answer is -2db
I tried doing (10dB)Log(1/100) but that equals -20dB
thanks
 A: Your answer of $-20$ dB is correct.
I suspect that you’re using a definition that gives the sound level in decibels as $10\log\left(\frac{I}{I_0}\right)$, where $I_0$ is the minimum perceptible intensity, or something very similar; if so, the following calculation shows exactly why $-20$ dB is correct.
Suppose that $I_A$ is the intensity of sound wave $A$, and $I_B$ is the intensity of sound wave $B$. Then the loudness of $A$ in decibels is $$L_A=10\log\left(\frac{I_A}{I_0}\right)\;,\tag{1}$$ and that of $B$ is $$L_B=10\log\left(\frac{I_B}{I_0}\right)\;.$$
You’re told that $I_A=\frac1{100}I_B$. Substituting that into $(1)$, we get
$$\begin{align*}
L_A&=10\log\left(\frac{I_A}{I_0}\right)\\
&=10\log\left(\frac{\frac1{100}I_B}{I_0}\right)\\
&=10\log\left(\frac{\frac{I_B}{I_0}}{100}\right)\\
&=10\left(\log\left(\frac{I_B}{I_0}\right)-\log 100\right)\\
&=10\log\left(\frac{I_B}{I_0}\right)-10\cdot2\\
&=L_B-20\;.
\end{align*}$$
A: Just 10*log10(1/100) =  -20 dB and you are done!
@Petersicecream:  you are right, multiply by 10. Thanks! 
