Number comparison: $5^{152}<2^{353}$ and $2^{1413}<3\cdot 5^{608}$

Question

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.

Answer 1

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 $\frac{2^7}{5^3}>1$.

Answer 2

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.

Answer 3

$$\begin{array} {} 5^{152} &\overset?\lt & 2^{353} \\ {5^2 \over 2^3} &\overset?\lt& {2^{350} \over 5^{150}} \\ {25 \over 8} &\overset?\lt& \left({2^7 \over 5^3}\right)^{50} \\ 3+{1\over 8} &\overset?\lt& \left(1+{3 \over 125}\right)^{50} & \qquad \text{ //binomial} & \text{expansion} \\ 3+{1\over 8} &\overset?\lt& 1+ 50 \cdot {3 \over 125} & + \binom{50}{2}\cdot {9 \over 125^2} &+ \cdots\\ 2+{1\over 8} &\overset?\lt& 0+ {6 \over 5} & + \binom{50}{2}\cdot {9 \over 125^2} &+ \cdots\\ 1-{3\over 40} &\overset?\lt& & + {50 \cdot 49 \over 2}\cdot {9 \over 125^2} &+ \cdots\\ 1-{3\over 40} &\overset?\lt& & 25 \cdot (50-1) \cdot {9 \over 25^3} &+ \cdots\\ {37\over 40} &\overset?\lt& & (50-1) \cdot {9 \over 25^2} &+ \cdots\\ {37\over 8} &\overset?\lt& & {18 \over 5} - {9 \over 5^3} &+ 5\cdot (\cdots)\\ 4+{5\over 8} &\overset?\lt& & 3+{3 \over 5} - {9 \over 5^3} &+ 5\cdot (\cdots)\\ 1+{5\over 8} &\overset?\lt& & {3 \over 5} - {9 \over 5^3} &+ 5\cdot (\cdots)\\ {65\over 40} &\overset?\lt& & {24 \over 40} - {9 \over 125} &+ 5\cdot ( \cdots) \\ {41\over 40} + {9 \over 125} &\overset?\lt& & &+ 5 \cdot \binom{50}{3}\cdot {27 \over 125^3} & + 5 \cdot(\cdots)\\ \vdots \end{array}$$ After this, my patience ran out, but only with that 4'th term of the binomial expansion the result would have been achieved, so this problem is surely solvable by paper&pen ...
Binomial expansion to 4 terms:
25/8 < 1 + 6/5 + 441/625 + 21168/78125 + (...) = 248168/78125 + (...)

Answer 4

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.