A simpler non-calculator proof for $17^{69}<10^{85}$

Question

I have proved that $17^{69}<10^{85}$ by using the following inequalities: $x<\exp\left(\dfrac{2(x-1)}{x+1}\right)$ for all $x\in \left]-1,1\right[$ and $x<{\mathrm e}^{x-1}$ for all $x\in \left] 1,+\infty \right[$, but I am looking for a simpler non-calculator proof.

My proof is the following: \begin{align*}\frac{17^{69}}{10^{85}}&=\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\left(\frac{5^3}{2^7}\right)^2\cdot\frac{5}{4}<\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\frac{5}{4}=\left(\frac{4913}{5000}\right)^{23}\cdot \frac{5}{4}\\&<\left(\exp\left(\frac{2\left(\frac{4913}{5000}-1\right)}{\frac{4913}{5000}+1}\right)\right)^{23}\cdot\exp\left(\frac{5}{4}-1\right)\\&=\exp\left(-\frac{174}{431}\right)\cdot\exp\left(\frac{1}{4}\right)=\exp\left(-\frac{265}{1724}\right)<1.\end{align*}

Could anyone find a simpler non-calculator proof without using big numbers?

Answer 1

I will also say some words on this. The general procedure to easily show such inequalities without computer is to... use the computer to get "close powers" of the bases, here $17$ and $10$, then use the coarsest that still does the job, and of course not mention that this was done so! In our case, i am forgetting in this second to not mention that the "first closest powers" of $17$ and $10$ come from the convergents of the continued fraction of $a=\displaystyle\log_{10} 17=\frac {\log 17}{\log 10}$, so let us show them...

sage: c = continued_fraction( log(17)/log(10) )
sage: cvgts = [ c.convergent(k) for k in [1..7] ]
sage: cvgts
[5/4, 11/9, 16/13, 283/230, 299/243, 1180/959, 1479/1202]

So we expect that

• $17^4=83521$ is "close" to $10^5$, yes, this is the case and $17^4\color{blue}{<}10^5$,
• $17^9=118587876497$ is "close" to $10^{11}$, yes, and $17^{9}\color{red}{>}10^{11}$,
• $17^{13}=9904578032905937$ is "close" to $10^{16}$, yes, and $17^{13}\color{blue}{<}10^{16}$,
• $17^{230}=\dots$ is "close" to $10^{283}$, yes, and $17^{230}\color{red}{>}10^{283}$, and so on.

Now we "completely forget" about the above, and write some inequalities. I will use the knowledge of the "steps" $17^4$, and $17^{13}$ below (of course, without mentioning this)... It will be a "hard job" (more than four lines) to establish $17^{13}\le 10^{16}$, but then we can relax and easily show the needed inequality. The most complicated operation will be to compute $836^2$ below. So let us start now!

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$$ \begin{aligned} 17^4 &= 289^2 = (290-1)^2 = 84100-580+1=83521\\ &< 83600\ , \\[3mm] % 17^{13} &=17\cdot (17^4)^3\\ &< 17\cdot 836^3\cdot 10^6 \\ &= 17\cdot 836^2\cdot 836\cdot 10^6 \\ &= 17\cdot 698896\cdot 836\cdot 10^6 \\ &< 17\cdot 700000\cdot 840\cdot 10^6 \\ &= 17\cdot 7\cdot 84\cdot 10^{12} \\ &= 9996\cdot 10 ^{12}\\ &< 10^{16}\ , \\[3mm] % 17^{69} &= (17^{13})^5\cdot 17^4\\ &<(10^{16})^5\cdot 83600\\ &< 10^{80}\cdot 10^5\\ &= 10^{85}\ . \end{aligned} $$

Answer 2

Claim 1: $2.3<\ln 10.$

Claim 2: $\ln 1.7<8/15$

Both these claims can be proven easily via Taylor series, etc.

Now, using the above inequalities, we have $1.7^{69}<e^{69\cdot \frac{8}{15}}<10^{16},$ or, multiplying $10^{69}$ on both sides, $17^{69}<10^{85}.$

Answer 3

You already got $$\frac{17^{69}}{10^{85}}\lt\left(\frac{4913}{5000}\right)^{23}\cdot \frac{5}{4}$$ from which we have $$\begin{align}\frac{17^{69}}{10^{85}}&\lt\left(\frac{4913}{5000}\right)^{23}\cdot \frac{5}{4} \\\\&\lt \left(\frac{4950}{5000}\right)^{23}\cdot \frac{5}{4} \\\\&=\left(\frac{99}{100}\right)^{23}\cdot \frac{5}{4} \\\\&=\bigg(1-\frac{1}{100}\bigg)^{23}\cdot \frac{5}{4} \\\\&=\frac 54\sum_{k=0}^{23}\underbrace{\binom{23}{k}\bigg(-\frac{1}{100}\bigg)^k}_{f(k)}\ \ \ \ \ \text{(binomial theorem)} \\\\&=\frac 54(f(0)+f(1)+\cdots +f(22)+\underbrace{f(23)}_{\lt 0}) \\\\&\lt \frac 54(f(0)+f(1)+\cdots +f(22)) \\\\&=\frac 54\bigg(f(0)+f(1)+f(2)+\sum_{k=1}^{10}(f(2k+1)+f(2k+2))\bigg) \\\\&=\frac 54\bigg(f(0)+f(1)+f(2) \\&\qquad+\sum_{k=1}^{10}\bigg(\binom{23}{2k+1}\bigg(-\frac{1}{100}\bigg)^{2k+1}+\binom{23}{2k+2}\bigg(-\frac{1}{100}\bigg)^{2k+2}\bigg)\bigg) \\\\&=\frac 54\bigg(f(0)+f(1)+f(2) \\&\qquad+\sum_{k=1}^{10}\bigg(\frac{-23!(\frac{1}{100})^{2k+1}}{(2k+1)!(23-2k-1)!}+\frac{23!(\frac{1}{100})^{2k+2}}{(2k+2)!(23-2k-2)!}\bigg)\bigg) \\\\&=\frac 54\bigg(f(0)+f(1)+f(2) \\&\qquad+\sum_{k=1}^{10}\frac{23!(\frac{1}{100})^{2k+2}}{(2k+2)!(22-2k)!}\bigg(-100(2k+2)+(22-2k)\bigg)\bigg) \\\\&=\frac 54\bigg(f(0)+f(1)+f(2)+\underbrace{\sum_{k=1}^{10}\frac{23!(\frac{1}{100})^{2k+2}(-202k-178)}{(2k+2)!(22-2k)!}}_{\lt 0}\bigg) \\\\&\lt\frac 54\bigg(f(0)+f(1)+f(2)\bigg) \\\\&=\frac 54\bigg(1-\frac{23}{100}+\frac{253}{10000}\bigg) \\\\&=\frac 54\cdot\frac{10000-2300+253}{10000} \\\\&=\frac{39765}{40000} \\\\&\lt 1\qquad\blacksquare\end{align}$$

Answer 4

Since $17^3 = 4913 < 492 × 10$, then$$ 17^6 < 492^2 × 10^2 = 242064 × 10^2 < 243000 × 10^2 = 3^5 × 10^5. $$ Now it suffices to prove that $(3^5 × 10^5)^{23} < (10^{85})^2$, or $3^{23} < 10^{11}$. Note that $3^9 = 27^3 = 19683 < 2 × 10^4$ and $3^5 = 243 < 25 × 10$, thus$$ 3^{23} = (3^9)^2 × 3^5 < (2 × 10^4)^2 × (25 × 10) = 10^{11}. $$

Answer 5

$$17 ^{ 13} = ((17^3)^2)^2 \cdot 17= (4913 \cdot 4913)^2\cdot 17< (242\cdot10^5)^2\cdot 17\\< 588\cdot10^{12}\cdot 17= 9996\cdot10^{12}<10^{16} $$

Hence, $$17 ^{ 69} = \left(17^{13}\right)^{\frac{69}{13}}<10^{16\cdot(5+\frac{4}{13})}= 10^{80+\frac{64}{13}} < 10^{85}.$$

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Here are some minor tricks to make the computation in the first inequalities even easier.

$$4913 \cdot4913=(4910+3)(4920-7) < 4910\cdot4920$$ $$491 \cdot492=(500-9)(500-8)= 241572$$ $$242\cdot242=(240+2)(245-3)<240\cdot245=12\cdot490= 58800$$

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Similar formulas:

$$\begin{array}{cl} \left.17^{4}\right/10^{5} &=0.83521\\ \left.17^{13}\right/10^{16} &=0.990458\!\cdots\\ \left.17^{69}\right/10^{85} &=0.796115\!\cdots\\ \left.17^{243}\right/10^{299} &=0.997902\!\cdots\\ \left.17^{1202}\right/10^{1479} &=0.999087\!\cdots\\ \left.17^{5524}\right/10^{6797} &=0.999636\!\cdots\\ \left.17^{7685}\right/10^{9456} &=0.999910\!\cdots\\ \vdots\\ \left.17^{302464054}\right/10^{372166569} &=0.99999999988\cdots\\ \end{array}$$

The above data is generated with, among other tools, the continued fraction of $$ \log_{17}10= 0.81271150929195899925562198972659\cdots,$$ which is, $$ [0; 1, 4, 2, 1, 17, 1, 3, 1, 1, 3, 3, 26, 1, 1, 2, 3, 2, 11, 64, 2, 3, 1, 13, 1, 8, 1, 4, \cdots].$$

Answer 6

Firstly, can be obtained the next numerical inequalities.

• $$5\cdot17^3 = 24565 < 24576 = 6\cdot 16^3,$$ $$\mathbf{\left(\dfrac{17}{16}\right)^3 <\dfrac65}.\tag1$$
• $$3^5\cdot2^6 =15552 < 15625 = 5^6,$$ $$\mathbf{3^5<\left(\dfrac52\right)^6}.\tag2$$
• $$(1.024)^4 < 1.0486^2 < 1 + 0.0972 + 0.0025 < 1.1,$$ $$1.1^7 = 1 + 0.7 + 0.21 + 0.035 + 0.0035 + 0.00021 + 0.000007 + 0.0000001 < 2,$$ $$2^{280} = (1.024)^{28}\cdot10^{84} < 2\cdot10^{84},$$ $$\mathbf{2^{279}<10^{84}}.\tag3$$

Then, taking in account $(1)-(3),$ one can get: $$\left(\dfrac{17}{16}\right)^{69} < \left(\dfrac65\right)^{23} = 27\cdot(3^5)^4\cdot\left(\dfrac25\right)^{23} < 27\left(\dfrac52\right)^{24}\left(\dfrac25\right)^{23} =\dfrac{135}2,$$ $$17^{69} < \dfrac{135}{2}\,\dfrac {2\cdot2^{279}}{16}<\dfrac{135}{16}\cdot10^{84},$$

$$\color{brown}{\mathbf{17^{69}< \dfrac{135}{16}\cdot10^{84}}},$$ $$\color{brown}{\mathbf{17^{69}<10^{85}.}}$$ Thus, there is a simple proof of a more rigorous inequality.

Answer 7

Proof that $17^{69} < 10^{85}$ without using calculator, there are many ways this can be done you'll need to understand number theory, so I'll highlight a few examples $$17^{69} < 10^{85}$$ $$17^{69} < 10^{17×5}$$ $$17^{1/17} < 10^{5/69}$$ Remember $$\lim_{n→∞} \sqrt[n]{n} = 1$$ Meaning the value of 17^{1/17} is between $1$ to $1.4$ $$17^{1/17} < 10^{5/69}$$ Using long division $\frac{5}{69} = 0.07246$ $$17^{1/17} < 10^{x}$$ But $10^x$ can also be between $1$ to $1.4$ if $x$ is from $0$ to $≈0.1$ So the the value $\frac{5}{69}$ is too small, therefore $$17^{69} \mathbb{<} 10^{85}$$

Another way $$17^{69} < 10^{85}$$ $$17^{69} < 10^{17×5}$$ $$17^{69/17} < 10^{5}$$ Using long division $\frac{69}{17} = 4.0588$ $$17^{4.0588} < 10^{5}$$ If we assume that $17^{4.0588} = 20^x$, since the base became bigger, the power would be smaller for them to be equal

$17 → 20$$, $4.0588 → ≈3$

$$17^{4.0588} ≈ 20^3$$ $$≈20^3 < 10^5$$ But we know clearly that $8000 < 10000$ Therefore $$17^{69} \mathbb{<} 10^{85}$$