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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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    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!!2011-06-29

4 Answers 4

5

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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    @Jack: Right, and D = 100 is the only answer in that range.2011-06-29
4

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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    Offtopic standard rant: "on computers a kilobyte is 1024" is not exactly universally true. I [wish](http://lpar.ath0.com/2008/07/15/si-unit-prefixes-a-plea-for-sanity/) this annoying abuse of notation would go away, but meanwhile see [w:Binary prefix](http://en.wikipedia.org/w/index.php?title=Binary_prefix&oldid=436463981).2011-06-29
2

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.

1

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 $n \approx 19.8$.