# Compare two powers of numbers without common divisor

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

• Not sure, but $2^{10}\approx 10^3$ and $3^2 \approx 10$ might be good approximations. – angryavian Jun 1 '15 at 19:32
• What about comparing $2^{15}$ and $3^{10}$? Since $3^{10}$ is bigger – Gregory Grant Jun 1 '15 at 19:33
• It is easier to compare these numbers using logarithms rather than their true values. So instead of using $m^n$ we can look at $\log m^n = n\log m$. – 3SAT Jun 1 '15 at 19:35
• and you could look up the values in a log table, so no calculator needed. – Brian Minton Jun 1 '15 at 20:46
• 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. – PM 2Ring Jun 2 '15 at 13:11

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

• 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... – CiaPan Jun 2 '15 at 5:59
• @CiaPan: It is very easy to calculate them even mentally! It is only four digits... – user21820 Jun 2 '15 at 8:48
• 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. – CiaPan Jun 2 '15 at 10:47
• @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. – Ethan Bolker Jun 2 '15 at 13:17
• @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 $3^5$. – CiaPan Jun 2 '15 at 13:41

We could also notice that

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

• This seems much more reliable than computing powers of three! – David Richerby Jun 1 '15 at 22:23

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: $3^{43} > 2^{60}$.

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

which is true by Bernoulli generalization:

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

• +1 clever, but not elementary enough for this OP. – Ethan Bolker Jun 2 '15 at 13:19

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

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

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

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 $2^{60}$.

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.

## protected by Zev ChonolesApr 6 '16 at 8:40

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?