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.
 A: We could also notice that
$3^{43} > 3^{40} = 9^{20} > 8^{20} = 2^{60}$.
A: 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, $3^{38} / 2^{60}  \approx 1.1716770936054666457995510$
A: Using $a^{bc} = (a^{b})^{c}$ we obtain:
$$3^{43} = 9^{43/2} > 9^{21} > 8^{20} = 2^{60}$$
A: 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 $2^{60}$.
A: 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.
A: Since $$3^7=2187\gt 1024=2^{10},$$
one has
$$3^{43}\gt 3^{42}=(3^7)^6\gt (2^{10})^6=2^{60}.$$
A: 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: $3^{43} > 2^{60}$. 
A: $$3^{43}>2^{60}\iff (3/2)^{43}>2^{17}\iff (1+0.5)^{\frac{43}{17}}>2,$$   
which is true by Bernoulli generalization:   
$$(1+0.5)^{\frac{43}{17}}\ge 1+0.5\cdot \frac{43}{17}>2$$
