Compare three numbers, expressed as powers: $4^{68}$, $5^{51}$ and $12^{23}$ So I have these numbers:
$4^{68}, 5^{51}, 12^{23}$
They need to be ordered from the smallest to greatest. I have no idea how to solve this. Typically, one should break down the exponents somehow to get the same base or exponent, but in this case it seems problematic as two of the exponents are prime numbers and the bases seem not to be expressable by each other.
Help would be appreciated.
 A: First note that $$12^{23}=(3\cdot4)^{23}=3^{23}\cdot4^{23}<4^{23}\cdot4^{23}=4^{46}\;,$$
so $12^{23}$ is definitely less than $4^{68}$. 
Next, $51$ and $68$ are both multiples of $17$, so let’s see whether that fact leads to anything nice. $4^{68}=(4^4)^{17}=256^{17}$, and $5^{51}=(5^3)^{17}=125^{17}$, so clearly $5^{51}<4^{68}$. It only remains to determine the relative sizes of $12^{23}$ and $5^{51}$. But that’s clear: $12^{23}<4^{46}<5^{51}$.
The final order is therefore $12^{23}<5^{51}<4^{68}$.
A: OK, assuming that we can't just take logs, I looked for numbers that were roughly similar in the power sequence of each base, just to see if the answer was clear:
$$ 4^{3.5}=128, 5^3=125, 12^2=144$$
seemed good enough for exploratory purposes. So then I'm comparing 
$$ 128^{19ish}, 125^{17}, 144^{11.5}$$
and it's clear that $4^{68}>5^{51}>12^{23}$.
A: Clearest way of all:
$$12^{23} = 6624737266949237011120128$$
$$5^{51} = 444089209850062616169452667236328125$$
$$4^{68} = 87112285931760246646623899502532662132736$$
A: Observe that $68=17 \times 4$ and $51 = 17 \times 3$, therefore $4^{68}=(4^4)^{17}$ and $5^{51} = (5^3)^{17}$. Compute the base, $4^4= 256$ is greater than $5^3 = 125$.
Then observe that $3 \times 23 = 69$, so $4^{68} = (4^3)^{23}/4 = 64^{23}/4$, which is much much much greater than $12^{23}$.
Therefore $12^{23}$ is the smallest.
Quantitatively, $5^{51} > 5^{46} = 25^{23} > 12^{23}$, which also shows that $12^{23}$ is the smallest.
Final answer:
$$12^{23} < 5^{51} < 4^{68}$$
