Number comparison: $5^{152}<2^{353}$ and $2^{1413}<3\cdot 5^{608}$ Is it possible to prove that $5^{152}<2^{353}$ and $2^{1413}<3\cdot 5^{608}$ without using a calculator or logarithms (middle school math only recommended)?
My idea for the first one was to use the obvious $5^3<2^7$ and then raise to the power of $50$ to get $5^{150}<2^{350}$. But since $5^2>2^3$, I couldn't use this approach to get the desired result.
Can you please help me find a relatively short proof for these inequalities? Thank you.
 A: As @Gottfried Helms pointed out, the binomial expansion works.
We have
\begin{align*}
 &5^{152} < 2^{353} \\
 \iff \quad & 10^{152} < 2^{505}\\
 \iff \quad & 10^{152} < 1024^{50}\cdot 2^5\\
 \iff \quad & 10^{152} < 10^{150}(1 + 3/125)^{50}\cdot 2^5\\
 \iff \quad & \frac{25}{8} < (1 + 3/125)^{50} \\
 \iff \quad & \frac{5}{2\sqrt{2}} < (1 + 3/125)^{25} \\
 \Longleftarrow \quad & \frac{5}{2\sqrt{2}} < 1 + 25 \cdot \frac{3}{125}
 + \frac{25\cdot 24}{2}\frac{3^2}{125^2}\\
 \iff \quad & \frac{5}{2\sqrt{2}} < \frac{1108}{625}
\end{align*}
which is true.
We are done.
A: Not an answer, but some thoughts too long to put in the comment section.
For $5^{152} < 2^{353}$, probably not, it's really tight. If you look at the continued fraction of $\ln(5)/\ln(2)$:
$$[2;3,9,2,2,4,\ldots]$$
The first approximation is $2+\frac 13 = \frac 73$ which gives you $5^3 < 2^7$. The next one is $2+\frac{1}{3+\frac 19}=\frac{65}{28}$ which yields $5^{28}> 2^{65}$. Or you can have $2+\frac{1}{3+\frac{1}{10}}=\frac{72}{31}$ and $5^{31} < 2^{72}$. So even if this is allowed it still doesn't work because
$$5^{152} = \frac{5^{155}}{5^3} < \frac{2^{360}}{5^3}=2^{353}\frac{2^7}{5^3}$$ but unfortunately $\frac{2^7}{5^3}>1$.
A: I don't think so because it is so close.  You can write
$$2^7=\left(1+\frac 3{125}\right)5^3\\
2^{350}=\left(1+\frac 3{125}\right)^{50}5^{150}$$
I asked Alpha for $\left(1+\frac 3{125}\right)^{50}$ and got about $3.2734$ so now you can write
$$2^{353} \approx 8 \cdot 3.2734 \cdot 5^{150}\\
\approx 26\cdot 5^{150}\gt 5^{152}$$
but you need to evaluate the $50^{th}$ power to within $4\%$.  I would say that is outside the reasonable range of hand computation.
