Comparing powers of 13 and 19

Question

There is a beginner olympiad question asking us to compare $13^{99}$ and $19^{93}$. This is easy enough, but I toyed around with it and got (via a calculator) that $13^{99}$ lies between $19^{86}$ and $19^{87}$. Just for fun, I tried proving both of them. I could prove that $19^{87}$>$13^{99}$, but couldn't prove the other one. Any proof for the same is appreciated.

Edit: The comments reflect some confusion. This is what I am asking for: Prove that $13^{99}$>$19^{86}$ .

Note: I am posting the proof I found for $19^{87}$>$13^{99}$. This is essentially $19^{29}$>$13^{33}$. From this we get that $(19/13)^{33}$>$19^{4}$. Now, $(19/13)^{3}$>$3$, so this reduces to $3^{11}$>$19^{4}$. Further manipulation gives $(27/19)^{4}$>$3$. Observe that $(27/19)^{2}$>$2$, so $(27/19)^{4}$>$4$>$3$. I know that this is not how proofs are presented, but this is just a gist. Moreover, the intention was to present a way in which the second proof should be structured- ideally, only clever mathematical manipulations. Not that any other method is bad, but I am looking for just basic maths. Another thought, $\dfrac {19^{87}}{13^{99}}$~$9.35$, and $\dfrac {13^{99}}{19^{86}}$~$2.03$, so good luck finding this one!

Answer 1

Here is a way to show $$ \tag{$\dagger$} 19^{86} < 10^{110} < 13^{99} $$ by not using some simple computer power. (Well, this does not mean that during the search of the proof such computer support was excluded.)

The right inequality is equivalent to $10^{10}<13^9$. We either use a simpler calculator, or - if really needed - with bare hands: $$ \begin{aligned} 13^9 &= (13^3)^3=2197^3\\ &>2190^3=(220-1)^3\cdot 10^3=(10648000-3\cdot48400 + 3\cdot 220-1)\cdot 10^3\\ &>2190^3=(220-1)^3\cdot 10^3=(10648000-3\cdot50000 )\cdot 10^3\\ &>10^7\cdot 10^3=10^{10}\ . \end{aligned} $$ The other side is slightly more complicated. (We want $19^{43}<10^{55}$ and note that $55/43$ is among the convergents of $\log 19\ /\ \log 10$. We will use the proximity of $19$ to $20$.) We have: $$ \begin{aligned} 19^{43} &= 20^{43} \cdot\left(\frac {19}{20}\right)^{43}\\ &= \left(\underbrace{2^7\cdot \left(\frac {19}{20}\right)^5}_{99.04396}\right)^8 \cdot \frac 1{2^{13}} \cdot\left(\frac {19}{20}\right)^3 \cdot 10^{43} \\ &<99.044^8 \cdot \frac 1{2^{13}} \cdot\left(\frac {19}{20}\right)^3 \cdot 10^{43} \\ &\overset{(?)} <10^{55}\ . \end{aligned} $$ The last inequality, the one marked with $(?)$ is then successively implied by... $$ \begin{aligned} 99.044^8 \cdot \frac 1{2^{13}} \cdot\left(\frac {19}{20}\right)^3 &\overset{(?)}< 10^{12} &&\text{would follow from...} \\ 99.044^8 \cdot 19^3 &\overset{(?)}< 10^{15}\cdot 2^{16} &&\text{which would follow from...} \\ 100^8 \cdot 19^3 &\overset{(?)}< 10^{15}\cdot 2^{16} \cdot 1.0096^8 &&\text{which would follow from...} \\ 68590 &\overset{(?)}< (4\cdot1.0096)^8 &&\text{which would follow from...} \\ 68590 &\overset{(?)}< 16.3^4 &&\text{which would follow from...} \\ 68590 &\overset{(?)}< 265^2 &&\text{which is...} \\ 68590 &< 70225 \ . \end{aligned} $$ And the last inequality is valid. So we go back throught the path of all $(?)$, and get the needed inequality $19^{43}<10^{55}$.

$\square$

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Here is a less crumpled way to get the inequality, this time not going through the rather close $19^{43}<10^{55}$, but using a refined version of $10^{10}<13^9$. Explicitly, i am using the computation $\color{blue}{13^9=2197^3=(2200-3)^3=10604499373}>1.06\cdot 10^{10}$, (or as in the first solution, get the estimate without getting explicitly the last digits,) inequality which is close enough at this one point to have less problems at "all other" points. Generally, in such cases solutions are accepted / are easier to find when higher powers are compared with powers of ten (times simple factors, or at least times factors that can be controlled in the estimation process).

We use first the rather generous estimation $19^5=2476099<2500000=\frac 14\cdot 10^7$. Then $$ \begin{aligned} 19^{86} &=19\cdot(19^5)^{17}\\ &<19\cdot\frac 1{2^{34}}\cdot 10^{119}\\ &=19\cdot\frac 1{2^{34}}\cdot 10^9\cdot 10^{110}\\ &<19\cdot\frac 1{2^{34}}\cdot 10^9\cdot \frac 1{1.06^{11}}\cdot 13^{99}\\ &=\frac{19}{16}\cdot\left(\frac {1000}{1024}\right)^3\cdot \frac 1{1.06^{10}}\cdot 13^{99}\\ &<\frac{19}{16}\cdot1\cdot \frac 1{1.12^5}\cdot 13^{99}\\ &<\frac{19}{16}\cdot \frac 1{1.75}\cdot 13^{99}\\ &=\frac {19}{28}\cdot 13^{99}\\ &<13^{99} \ , \end{aligned} $$ and i hope there is no error on the above path.

Answer 2

I don't know if self-answers are an accepted norm, but I came up with a solution:

I will be using that $\frac{19^{3}}{13^{3}}$ is approximately $3.1219$.

Now, squaring both sides, I get that $\frac{19^{6}}{13^{6}}$<$10$. Perform some rearrangements to get $\frac{13^{6}}{19^{5}}$>$1.9$.

Raise both sides to the fourth power $\frac{13^{24}}{19^{20}}$>$1.9^{4}$=$13.0321$>$13$.

We get that $13^{23}$>$19^{20}$. Again raise to the fourth power to obtain $13^{92}$>$19^{80}$.

A small diversion: We had proved that $\frac{19^{6}}{13^{6}}$<$10$, so this is also less than 13, or $13^{7}$>$19^{6}$.

Finally, multiply this with the previously obtained result to get $13^{99}$>$19^{86}$.