Common logarithm word problem Each time a ray of light passes through a glass plate, it loses $\frac{1}{10}$ of its intensity. How many pieces of similar glass plates are needed to make the light intensity less than $\frac{1}{3}$ of its original value?
Let $x$ be the original intensity value of the ray of light.
Let $n$ be the number of needed similar glass plates to make the light intensity less than $\frac{1}{3}$ of its original value.
Thus
$x(\frac{1}{10})^n = (\frac{2}{3})x$
$(\frac{1}{10})^n = (\frac{2}{3})$
$log_{10} {10^-n} = log_{10} {2} – log_{10} {3}$
$n × log_{10} {10} = log_{10} {2} – log_{10} {3}$
Since $log_{10}{2} ≈ 0.3010$ and $log_{10}{3} ≈ 0.4771$, then
$n ≈ 0.3010 – 0.4771$
$n ≈ 0.3010 – 0.4771$
$n ≈ -0.1761$
I got a negative answer for $n$. This is where I got stuck. Any comments and suggestions will be much appreciated. Thank you in advance.
 A: The ray loses $\frac{1}{10}$ of its intensity. This means that the intensity of the ray after it leaves the glass plate is $\frac{9}{10}x$ (assuming that $x$ is the original intensity).
A: The intensity transmitted by a single plate is $0.9$, and by $n$ plates is $0.9^n$. If you want the smallest $n$ such that this value not larger than $0.3333$,
$$0.9^n\le0.3333$$
then
$$n\log0.9\le \log0.3333$$
$$n=\left\lceil\frac{\log0.3333}{\log0.9}\right\rceil=11.$$
Check:
$$0.9^{10}=0.3487\cdots,\\0.9^{11}=0.3138\cdots$$

Notes:

*

*The basis of the logarithm does not matter, as you are taking a ratio.


*As the ratio is not close to an integer, using the approximation $0.3333$ is fine.
A: After passing through a glass, remaining intensity $= \frac{9x}{10}$ if the initial intensity is $x$.
After passing through $n$ glasses, the remaining intensity $= (\frac{9}{10})^n \times x \lt \frac{x}{3}$
A: You can also just use some basic calculus-
$$\frac{dx}{dn}=\frac{x}{10}$$
$$\Rightarrow \int_{x}^{\frac{x}{3}}\frac{1}{x}dx=-\int_{0}^{n}\frac{1}{10}dn$$
$$\Rightarrow \ln{\frac{1}{3}}=-\frac{n}{10}$$
$$\Rightarrow \ln{3}=\frac{n}{10}$$
$$\Rightarrow n≈10.981…$$
As we need $n$ to be a natural number, $\fbox{n≥11}$
