15
$\begingroup$

Which of the numbers $2^{60}$ and $3^{43}$ is greater? There is no common divisor and it must be done without a calculator.

$\endgroup$
5
  • 2
    $\begingroup$ Not sure, but $2^{10}\approx 10^3$ and $3^2 \approx 10$ might be good approximations. $\endgroup$
    – angryavian
    Jun 1 '15 at 19:32
  • 1
    $\begingroup$ What about comparing $2^{15}$ and $3^{10}$? Since $3^{10}$ is bigger $\endgroup$ Jun 1 '15 at 19:33
  • 4
    $\begingroup$ It is easier to compare these numbers using logarithms rather than their true values. So instead of using $m^n$ we can look at $\log m^n = n\log m$. $\endgroup$
    – 3SAT
    Jun 1 '15 at 19:35
  • 1
    $\begingroup$ and you could look up the values in a log table, so no calculator needed. $\endgroup$ Jun 1 '15 at 20:46
  • 4
    $\begingroup$ Back in the day, many regular users of log tables wouldn't even need to look up the base 10 logs of 2 and 3 - they'd have them memorized to (at least) 4 significant figures. $\endgroup$
    – PM 2Ring
    Jun 2 '15 at 13:11
78
$\begingroup$

We could also notice that

$3^{43} > 3^{40} = 9^{20} > 8^{20} = 2^{60}$.

$\endgroup$
1
  • 5
    $\begingroup$ This seems much more reliable than computing powers of three! $\endgroup$ Jun 1 '15 at 22:23
23
$\begingroup$

Since $$3^7=2187\gt 1024=2^{10},$$ one has $$3^{43}\gt 3^{42}=(3^7)^6\gt (2^{10})^6=2^{60}.$$

$\endgroup$
5
  • 1
    $\begingroup$ How do you know $3^7 =2187$ and $1024=2^10$ without a calculator? Some people can remember these values but certainly not everybody, and I doubt one could calculate them with a paper and pencil during an exam... $\endgroup$
    – CiaPan
    Jun 2 '15 at 5:59
  • 21
    $\begingroup$ @CiaPan: It is very easy to calculate them even mentally! It is only four digits... $\endgroup$
    – user21820
    Jun 2 '15 at 8:48
  • $\begingroup$ Depends on your practice... Anyway I feel 'without a calculator' means 'with as little explicit calculations as possible' — some people can just memorize all powers of 2 up to $2^{16}$ (say, computing fans) while others need a calculator to evaluate $3^5$. Forcing a verbatim meaning of 'use of calculator' prevents those 'others' from finding $3^7<2^{10}$ the way mathlove did it. $\endgroup$
    – CiaPan
    Jun 2 '15 at 10:47
  • 2
    $\begingroup$ @CiaPan $2^{10} = 1024 \approx 1000 = 10^3$ is worth remembering. It makes many approximations easy. It's also why a kilobyte is 1024 bytes even though :"kilo" means "multiply by 1000". I agree that the value of $3^7$ is not nearly as useful. $\endgroup$ Jun 2 '15 at 13:17
  • $\begingroup$ @EthanBolker Yes, I know. I also know how much $2^{16}$ is. But I also know people, who know by heart logarithm base $e$, astronomical unit in kilometers, number of grads in one radian or the copper conductivity to the fourth decimal place, but have no idea how much $2^8$ is. They can easily calculate it, but do not know it, because they never needed it. That's why I asked my question, how one can solve the problem without that much calculation if one does not know $2^{10}$ or $3^5$. $\endgroup$
    – CiaPan
    Jun 2 '15 at 13:41
15
$\begingroup$

If we look at the powers of 3: 3, 9, 27, 81, 243, 729, 2187. 2187 looks pretty close to a power of 2: 2048. So let's start with that:

$$3^7 > 2^{11}$$

Take both to the 5th power:

$$3^{35} > 2^{55}$$

Obviously:

$$3^8 > 2^5$$

Multiplying those together: $3^{43} > 2^{60}$.

$\endgroup$
12
$\begingroup$

$$3^{43}>2^{60}\iff (3/2)^{43}>2^{17}\iff (1+0.5)^{\frac{43}{17}}>2,$$

which is true by Bernoulli generalization:

$$(1+0.5)^{\frac{43}{17}}\ge 1+0.5\cdot \frac{43}{17}>2$$

$\endgroup$
1
  • $\begingroup$ +1 clever, but not elementary enough for this OP. $\endgroup$ Jun 2 '15 at 13:19
7
$\begingroup$

Musically inclined mathematicians should know that an interval of 12 equally-tempered perfect fifths in just intonation is slightly larger than 7 octaves, i.e.,

$(3/2)^{12} > 2^7.\ $ FWIW, $(3/2)^{12} \approx 129.746$

In the standard 12 tone equally tempered scale, 12 perfect 5ths is exactly 7 octaves, i.e., a perfect 5th is 7 semitones. 12 semitones make one octave, so an equally-tempered semitone is a frequency ratio of $2^{1/12}$, and an equally-tempered perfect 5th is $2^{7/12} \approx 1.4983$, i.e., it's slightly flat compared to the "pure" perfect 5th of just intonation.

$$\begin{align} (3/2)^{12} & > 2^7\\ 3^{12} & > 2^{19}\\ 3^{36} & > 2^{57}\\ 9 * 3^{36} > 8 * 3^{36} & > 8 * 2^{57}\\ 3^{38} & > 2^{60}\\ \end{align}$$

which gives us somewhat tighter bounds than the OP. :)

FWIW, $3^{38} / 2^{60} \approx 1.1716770936054666457995510$

$\endgroup$
6
$\begingroup$

Using $a^{bc} = (a^{b})^{c}$ we obtain:

$$3^{43} = 9^{43/2} > 9^{21} > 8^{20} = 2^{60}$$

$\endgroup$
3
$\begingroup$

We have $3^{40}=81^{10}$. This is somewhat bigger than $2^{30}10^{10}$, which is somewhat bigger than $10^{19}$. And $3^{43}=27\times 3^{40}$.

Using $2^{10}\approx 1000$, we find that $2^{60}$ is of size about $10^{18}$. So it's not even close, $3^{43}$ is about $270$ times $2^{60}$.

$\endgroup$
3
$\begingroup$

Yet another way (besides looking for approximate values with exponents) is using log tables:

Comparing $2^{60}$ to $3^{43}$ is equivalent to comparing $60 \log(2)$ to $43 \log(3)$ [from laws of indices]

From tables: $\log_{10}(2) \approx 0.3$, $\log_{10}(3) \approx 0.477$

Then $60 \times \log(2) \approx 18.0$ ...

...but $43 \times \log(3) \approx 43 \times 0.477 \approx 20.5$

Hence $3^{43}$ is bigger; and by about $2.5$ in $\log_{10}$ land, which corresponds to a factor of $10^{2.5} \approx 300$ in numbers.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.