Common logarithm word problem

Question

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

$n ≈ 0.3010 – 0.4771$

$n ≈ -0.1761$

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.

score 2 · Answer 1 · answered Sep 29 '20 at 08:39

2

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).

answered Sep 29 '20 at 08:39

5xum

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Thank you for your comment. Should I say ${\frac{9}{10}}^{n}x={\frac{2}{3}}x$? – AYA Sep 29 '20 at 08:51
@MAK yes, that is the correct equation. – Crostul Sep 29 '20 at 09:24

score 1 · Answer 2 · answered Sep 29 '20 at 09:24

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.

score 0 · Answer 3 · answered Sep 29 '20 at 09:14

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After passing through a glass, remaining intensity $= \frac{9x}{10}$ if the initial intensity is $x$.

After passing through $n$ glasses, the remaining intensity $= (\frac{9}{10})^n \times x \lt \frac{x}{3}$

answered Sep 29 '20 at 09:14

Math Lover

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score 0 · Answer 4 · answered Sep 29 '20 at 11:35

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, $\fbox{n≥11}$

Common logarithm word problem

4 Answers4