Comparing two large numbers Can you compare two large exponential numbers, like $5^{44}$ and $4^{53}$ without taking their logs?
 A: $$5^{44}<5^{45}=(5^3)^{15}=125^{15}<128^{15}=(256/2)^{15}=4^{60}/2^{15}<4^{53}$$ (because $2^{15}>2^{14}=4^{7}$)
A: Using a GCD-like approach, start by dividing out the "smaller" term:
$${5^{44}\over 4^{44}}=\left(\frac54\right)^{44}\\
{4^{53}\over 4^{44}}=4^9$$
Now the new "smaller" term is $\frac54$:
$${\left(\frac54\right)^{44}\over \left(\frac54\right)^{18}}=\left(\frac54\right)^{26}\\
{4^9\over \left(\frac54\right)^{18}}=\left(\frac85\right)^{18}\\
\left(\frac54\right)^{8}\text{ vs }\left(\frac{32}{25}\right)^{18}$$
By inspection, $\frac{32}{25}\ge\frac54$, and as the exponent on $\frac{32}{25}$ is also greater, it is the greater quantity.  It is from the $4^{53}$ term, and therefore this is the greater original quantity.
A: One approach is to figure that, roughly, $2^{10} \approx 10^3$, and $5^9 \approx 2,000,000 = 2 \cdot 10^6$. 
Then, 
\begin{align}
4^{53} &= (2^2)^{53} \\
&= 2^{106} \\
&= 2^{100} \cdot 2^6 \\
&= (2^{10})^{10} \cdot 2^6 \\
&\approx (10^3)^{10} \cdot 2^6 \\
&= 10^{30} \cdot 2^6
\end{align}
By way of comparison,
\begin{align}
5^{44} &= 5^{45} \div 5 \\
&= (5^9)^5 \div 5 \\
&\approx (2 \cdot 10^6)^5 \div 5 \\
&= 2^5 \cdot (10^6)^5 \div 5 \\
&= 10^{30} \cdot (2^5 \div 5).
\end{align}
Now, $2^6 > 2^5 \div 5$, so one might suppose $4^{53} > 5^{44}$.
A: I'll interpret the actual question as "without using a calculator?" since I assume that's your objection to using logs. 
What I know about $5$ and $4$ is that $5^3 = 125$ is pretty close to but a bit smaller than $2^7 = 128$. That tells me that $\log_2 5 \le\frac{7}{3}$, hence that
$$\log_2 5^{44} \le \frac{308}{3} < 103$$
while $\log_2 4^{53} = 106$. So $4^{53}$ is bigger. Of course I could rephrase this argument without using logs by exponentiating everything but what's the point? 
A calculator will verify that $\frac{7}{3}$ is in fact a convergent of $\log_2 5$, whose continued fraction approximation begins $2 + \frac{1}{3 + \frac{1}{9 + \dots}}$. So the above is a pretty close approximation; in fact the true value of $\log_2 5^{44}$ is $102.164 \dots$. 
