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The question is motivated from the following problem:

At a $15$ percent annual inflation rate, the value of the dollar would decrease by approximately one-half every $5$ years. At this inflation rate, in approximately how many years would the dollar be worth $\frac{1}{1,000,000}$ of its present value?

$A.25\quad B.50\quad C.75\quad D.100\quad E.125$

Finally one may get $10^6=2^n$ and $5n$ is the desired approximation. With a step of calculation, $$n=\frac{6\ln10}{\ln 2}.$$ I have no idea how to go on without a calculator. Are there any tricks/techniques for solving the problem?

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Surely you know a power of 2 which is nearly a power of 10 and which enables you to get a good estimate of n without taking logs? Or you can find one quite quickly, or even - the long way - write down the doubling sequence starting 1, 2, 4, 8 ... even up to a million doesn't take long. – Mark Bennet Jun 29 '11 at 15:01
Specifically, noting that $2^{10} = 1024 \approx 1000$, we have $2^{20} = (2^{10})^2 \approx 1000^2 = 10^6$. – Shai Covo Jun 29 '11 at 15:11
Offtopic: As most software still annoyingly use, 1 megabyte = $2^{20}$ bytes, but mega is actually the SI prefix for $10^6$. This may help someone remember that $2^{20} \approx 10^6$. – ShreevatsaR Jun 29 '11 at 15:19
The telecommunications engineers know that doubling the signal power improves the signal-to-noise-ratio by 3dB. Or as Shai and Mark told you, $log_{10}2\approx 0.3$. Use base ten!! – Jyrki Lahtonen Jun 29 '11 at 15:20
up vote 5 down vote accepted

Using the change of base formula, we find that $\frac{ln10}{ln2} = log_2(10)$, which is between 3 and 4... so $5n$ is 30 times this.

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Hmm, this is good enough to get $90<5n<120$. – Jack Jun 29 '11 at 15:11
@Jack: Right, and D = 100 is the only answer in that range. – The Chaz 2.0 Jun 29 '11 at 15:23

Since $6 \ln 10 = \ln(10^6)$, the question asks: $10^6$ is approximately what power of 2?

You probably know that in science, "kilo" means 1000, but on computers a kilobyte is 1024. Why the strange number? Because it is a power of 2, and computers like powers of 2. So remember this: $10^3 \approx 2^{10}$. For your problem, square to get $10^6 \approx 2^{20}$. So your answer is approximately $20$.

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Well, $n$ is approximately 20 ... – The Chaz 2.0 Jun 29 '11 at 15:39
OK, the answer to the title is approximately 20. The answer to the "following problem" is $5\times 20$. – GEdgar Jun 29 '11 at 15:42
Offtopic standard rant: "on computers a kilobyte is 1024" is not exactly universally true. I wish this annoying abuse of notation would go away, but meanwhile see w:Binary prefix‌​. – ShreevatsaR Jun 29 '11 at 20:53

Note that $\ln 10 = \ln 5 + \ln 2$. Therefore changing the base to 2,

$n = \frac{6 (\ln 5 + \ln 2)}{\ln 2} = 6 \log_{2}5 + 6$

Since $\log_{2}5$ is slightly more than 2, we could estimate the total to be somewhere around 19.

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Looking at $10^6 = 2^n$ you can use the fact that $2^3 = 8$ and $2^4 = 16$ to approximate $10 \approx 2^{3.3}$ (or something nearer 3 than 4), with this you have that $(2^{3.3})^6 \approx 2^n$, so $n \approx 19.8$.

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