Formula for transparency! A transparent a plastic absorbs per cm $30\%$ of the light falling through it. How thick must the plastic be to absorb $60\%$ of the light?
 A: After $1$ cm of substance the light is $70$% or $0.7$ of what it was. After another $1$cm it's reduced again, to $0.7 \cdot 0.7=0.7^2=0.49$ of the initial value. So after $n$ cm it's reduced to 
$0.7^n$ of the initial value. Now you want to reduce with $60$%, that is, to $40$% or $0.4$ of the value. So you want $n$ so that
$$0.7^n = 0.4$$
The answer is 
$$n = \log_{0.7}0.4$$
This can also be calculated as 
$$\log_{0.7}0.4 = \frac{\log 0.4}{\log 0.7}$$ where the logs are in any base you want. 
Get
$$\frac{\log 0.4}{\log 0.7}= 2.56...$$
So you need about $2.5$ cm. Notice that $0.7^3=0.343$, $0.7^2=0.49$ and $0.7^{2.56..}=0.4$
Obs: 
In base $e$ $$\log 0.4 = \log (1-0.6) = - (0.6 + \frac{0.6^2}{2} + \frac{0.6^3}{3} + \ldots )\\
\log 0.7 = \log (1-0.3) = - (0.3 + \frac{0.3^2}{2} + \frac{0.3^3}{3} + \ldots )$$
and $\frac{\log 0.4}{\log 0.7} = \ldots$ can be calculated approximately without a calculator. 
A: Each centimeter transmits $70\%$ of the light that falls on it. Therefore, the formula would be
$$
p_{\text{absorbed}}=1-e^{\log(.7)\frac{w}{1\text{ cm}}}
$$
Solving for $p_{\text{absorbed}}=.6$ gives $w=\frac{\log(.4)}{\log(.7)}=2.569$ cm.
