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This is problem 19.d from Serge Lang's Basic Mathematics. Before ranting about my troubles, the description given is the following:

  1. You are testing the effect of a noxious substance on bacteria. Every 10 min, one-tenth of the bacteria which are still alive are killed. If the population of bacteria stats with $10^6$, how many bacteria are left after:

d. Within which period of $10 \text{ min}$ will $70\text{%}$ of the bacteria be killed?

[Note: If one-tenth of those alive are killed, then nine-tenths remain.]

Now I can solve this in a rote fashion, by dividing $10^6$ by $3$ to achieve a value that gives me bounds to put around it after rotely dividing $10^6$ by one-tenth and subtracting it from the dividend to reach a difference that will signify $10$ minutes have passed and taking this difference and dividing it by $\frac{1}{10}$ and subtracting it from the dividend [. . .] etcetera.

Because of the hint given for this problem, I was wondering if there was a less rote way to do this, though. Supposedly this book is full of insightful basic mathematics problems, so I was hoping to discover a different way of going about this. Is there?

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We start with this:

$$p(0)=10^6$$

After 10 minutes,

$$p(10)=10^6\times\frac9{10}$$

After 20 minutes,

$$p(20)=10^6\times\left(\frac9{10}\right)^2$$

After $10n$ minutes,

$$p(10n)=10^6\times\left(\frac9{10}\right)^n$$

We want this equal to $10^6\times\frac3{10}$, so

$$10^6\times\frac3{10}=10^6\times\left(\frac9{10}\right)^n$$

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the percentage of bacteria left after $10n$ minutes is $\frac{9}{10}^n$ so you need to find the smallest integer such that $\frac{9}{10}^n \leq 0.3$.

In other words $\lceil \log_{9/10} 0.3\rceil = \lceil \ln 0.3 / ( \ln(9)-\ln(10) )\rceil=\lceil 11.42 \rceil =12$, hence we want $120$ minutes, AKA $2$ hours.

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