人民日报看黑龙江2018--黑龙江频道--人民网

Question

1. You may not use a calculator or computer.

2. You may write "ln(X)" or "log(X)" to indicate the natural logarithm of X. Else, please let the reader know "log(X)" means log of X to the base 10, just to mention another common base.

3. You are allowed to use $\ln(1 + x)\approx\ x-\dfrac{x^2}{2} + \dfrac{x^3}{3} \ $ for appropriate small values of $x$.

4. To reduce some arithmetic, you are allowed to use these if they were to come up in calculations:

   $\ln(2) \ \approx \ 0.6931 $

   $\ln(3) \ \approx \ 1.0986 $

   $\ln(5) \ \approx \ 1.6094 $

5. If it were to come up, you may use $\ \dfrac{\ln(5)}{\ln(4)} \ \approx \ 1.161$ in a calculation.

---

Please show the steps without using a calculator or computer to indicate which expression is larger:

$\large4^{5^9} \ \ \ \text{or} \ \ \ 5^{6^8}$

Answer 1

Working by hand, I can show that

$$\begin{matrix}5^1 & 5 & 6^1 & 6 \\5^2 & 25 & 6^2 & 36 \\5^3 & 125 & 6^3 & 216 \\5^4 & 625 & 6^4 & 1296 \\5^5 & 3125 & 6^5 & 7776 \\5^6 & 15625 & 6^6 & 46656 \\5^7 & 78125 & 6^7 & 279936 \\5^8 & 390625 & 6^8 & 1679616 \\5^9 & 1953125 & & \end{matrix}$$

So $5^9 = 1953125$ and $6^8 = 1679616$

Compare

$$4 ^ {5 ^ 9} \text{ vs } 5 ^ {6 ^ 8} $$ $$4^{1953125} \text{ vs } 5^{1679616}$$ $$ 1953125\times\ln{4} \text{ vs } 1679616\times\ln 5$$ $$\frac{1953125}{1679616} \text{ vs } \frac{\ln{5}}{\ln{4}}$$

Divide longhand to show that this is true

The longhand division shows $$\frac{1953125}{1679616} > 1.162$$ and, when knowing $\frac{\ln{5}}{\ln{4}}$ is 1.161 to 3 decimal places, means $$\frac{1953125}{1679616} > \frac{\ln{5}}{\ln{4}}$$ and thus $$4 ^ {5 ^ 9} > 5 ^ {6 ^ 8} $$

Answer 2

The claim to check is

$$4^{5^9} \stackrel{?}{>} 5^{6^8}$$

We will check it without logarithms.

Take the $5^8$th root of both sides to get the equivalent:
$$4^{5} \stackrel{?}{>} 5^{1.2^8}$$
We can calculate ${1.2^8}$ by hand by squaring 3 times: $1.2 \to 1.44 \to 2.0736 \to 4.29981696$. This is just below $4.3$

So it is enough to check

$$4^{5} \stackrel{?}{>} 5^{4.3} (> 5^{1.2^8})$$ which is equivalent to $$4^{50} \stackrel{?}{>} 5^{43}$$ which is equivalent to $$2^{143} \stackrel{?}{>} 10^{43}$$ which is equivalent to $$(2^{10})^{14} \cdot 2^3 \stackrel{?}{>} (10^3)^{14} \cdot 10 $$

Since $2^{10}=1024$, it is equivalent to

$$1.024^{14}\cdot 8 \stackrel{?}{>} 10$$ which is true because $(1+x)^n > 1+nx$ gives $$1.024^{14} > 1+14\times 0.024 = 1.336 > 1.25$$

So,

$$4^{5^9} > 5^{6^8}.$$

Answer 3

Starting with the claim

$$ 4^{5^9} > 5^{6^8} $$

We have

$$ \iff 5^9\ln4 > 6^8\ln5 $$ $$ \iff \frac{5^9}{6^8} > \frac{\ln5}{\ln4} $$ $$ \iff 5\times\left(\frac56\right)^8 > \frac{\ln5}{\ln4} $$ $$ \iff \ln5+8\ln\frac56 > \ln\frac{\ln5}{\ln4} $$

Using $\ln x = - \ln(1/x)$, we can define a quantity $A$:

$$ \iff A:= \frac{\ln5 - \ln\frac{\ln5}{\ln4}}{\ln\frac65} > 8 $$

Using the hints we have the bounds

$$ 0.69305 < \ln(2) < 0.69315$$ $$ 1.09855 < \ln(3) < 1.09865$$ $$ 1.60935 < \ln(5) < 1.60945$$

Since we want to prove that $A>8$, we have to try to make it smaller using our bounds while still having the result be larger than $8$.

For the first term in the numerator, if we make it smaller, then $A$ will get smaller, so we need $$\ln(5) > 1.60935.$$ For the second term in the numerator, if we make it bigger, then $A$ will get smaller, so we can first do longhand division to get $$ \frac{\ln5}{\ln4} = \frac{\ln5}{2\times\ln2}< \frac{1.60945}{2\times 0.69305} < 1.1612 $$ (the given hint only upper bounds it by 1.1615 which will not be good enough), then put this in the series to compute $$\ln\frac{\ln5}{\ln4} < \ln 1.1612 < 0.1612 - \frac{0.1612^2}{2} + \frac{0.1612^3}{3}$$ (it is an upper bound since we have just added a positive term, the cubed term, to an alternating series). We can get an upper bound on this by only ever keeping 4 decimal places. Let $x=0.1612$ then compute $0.0258 < x^2 < 0.0260$. Use the upper bound for $x^2$ to compute $x^3 < 0.0260x < 0.0042$, so this is $$<0.1612 - 0.0129 + 0.0014 = 0.1497$$ For the denominator, if we make it bigger, then $A$ will get smaller, so we need $$ \ln(6/5) = \ln2 + \ln3 - \ln5 < 0.69315 + 1.09865 - 1.60935 = 0.18245 $$

we get that the claim

$$ \iff \frac{\ln5 - \ln\frac{\ln5}{\ln4}}{\ln(6/5)} > \frac{1.60935 - 0.1497}{0.18245} = 8 + \frac{1}{3649}> 8$$

which is true

Answer 4

Let us first try to reduce the exponents to manageable sizes. The idea is to find positive integers $x$ and $y$ such that $4^{5^9 - x} > 5^{6^8 + y}$ while $\text{gcd}(5^9 - x, 6^8 + y)$ is large. We can find that the largest gcd is $54253$, achieved when $x = 17$ and $y = 2227$. Now it suffices to show that $$ 4^{5^9 - 17} > 5^{6^8 + 2227} \Leftrightarrow 4^{54253 \cdot 36} > 5^{54253 \cdot 31} \Leftrightarrow 4^{36} > 5^{31}$$ The numbers are a quite big but it's possible to compute by hand: \begin{align*} 4^{36} &= 4\,722\,366\,482\,869\,645\,213\,696 \\ 5^{31} &= 4\,656\,612\,873\,077\,392\,578\,125 \end{align*} From here we can see that $4^{36} > 5^{31}$ and therefore $4^{5^9} > 5^{6^8}$.

Answer 5

For starters,

We wish to show that $4^5 > 5^{1.2^8}$, as raising both sides to the power $5^8$ will resolve the initial inequality, so we will need to prove that $\frac{5\ln(4)}{\ln(5)} > 4.306 > 1.2^8$.
The first inequality follows from the given precisions for $\ln(2)$ and $\ln(5)$, even if we were given the worst combination of rounding errors. Simply using $\frac{5}{1.161}$ would give us $4.3066$ to work with instead.

Time to dust off an old mathematical trick!

Using the series for $\ln(1+x)$ whose first few terms have been given to us, subtract from it a copy with $x$ negated to yield $\ln\left(\frac{1+x}{1-x}\right)= 2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\dots\right)$. Substituting $\frac{1}{2n+1}$ yields a particularly useful form:
$1.) \ln\left(1+\frac{1}{n}\right) = \frac{2}{2n+1}+\frac{2}{3(2n+1)^3}+\frac{2}{5(2n+1)^5}+\dots$
We can get a strong upper bound by replacing coefficients higher than 3 with 3, and using the standard rule of geometric series sums to get
$2.) \ln\left(1+\frac{1}{n}\right) < \frac{2}{2n+1}+\frac{1}{6n(n+1)(2n+1)}$

We can apply this rule two different ways to show the final inequality:

Using 2.) with $n=5$ yields $8\ln(1.2) < \frac{16}{11}+\frac{2}{495} < 1.4586$.
Using 1.) with a single term, we have that $\ln(4.306) = \ln(4)+\ln(1.0765) > 1.3862+\ln\left(1+\frac{1}{13.1}\right) > 1.3862+\frac{1}{13.6}> 1.4597$.
And we are done!

Answer 6

This answer begins similarly to ThomasL's, but diverges partway through.

We want to know if $4^{5^9} > 5^{6^8}$. That's the same as asking whether $$ \ln{4^{5^9}} \stackrel{?}{>} \ln{5^{6^8}}$$ $$ 5^9 \ln{4} \stackrel{?}{>} 6^8 \ln{5} $$ or $$ \frac{5^9}{6^8} \frac{\ln{4}}{\ln{5}} \stackrel{?}{>} 1 $$

Since we are given

$$ \frac{\ln{5}}{\ln{4}} \approx 1.161 $$

we need only determine whether

$$ \frac{5^9}{6^8} \stackrel{?}{>} 1.161 $$

Now I can compute by hand

$$ \begin{aligned} \ln{\frac{5^9}{6^8}} & = 9 \ln 5 - 8\left(\ln 3 + \ln 2\right) \\ & = 9 \times 1.6094 - 8 \times (1.0986 + 0.6931) \\ & = 14.4846 - 14.3336 \\ & = 0.1510 \end{aligned}$$ If this were negative, I'd be able to stop, because then we'd know $5^9 < 6^8$, and its product with $\ln 4 / \ln 5$ would be smaller than unity.

However, I can make use of the fact that

$$ e^x = 1 + x + \frac{x^2}{2} + ... $$

to work out that

$$ \begin{aligned} \frac{5^9}{6^8} \approx e^{0.1510} & \approx 1 + 0.1510 + 0.5 \times (0.1510)^2 + \ldots \\ & = 1.1510 + 0.5 \times 0.022801 + \ldots \\ & = 1.1624 + \ldots \end{aligned}$$ This is already larger than the 1.161 I'm comparing it to, so I don't need to compute any more terms in the expansion; they're all positive so they will only increase the sum.

Thus

$$ \frac{5^9}{6^8} \frac{\ln 4}{\ln 5} > 1 $$

and so

$$ 4^{5^9} > 5^{6^8} $$

Answer 7

Originally I wanted to avoid relying on any of the logarithm approximations, unfortunately the results are just close enough that we must rely on one of them. This solution seeks to use methods that make by-hand calculations easier to reason about, due to seeing some comments saying it's much too hard to calculate by hand.

$$4^{5^9} \stackrel{?}{>} 5^{6^8}$$ $$5 \cdot 5^8 \cdot 2 \ln(2) \stackrel{?}{>} 6^8 \cdot \ln(5)$$ $$10 \frac{\ln(2)}{\ln(5)} \stackrel{?}{>} \left ( \frac{6}{5} \right )^8$$ $$\frac{\ln(2)}{\ln(5)} \stackrel{?}{>} 10^{-1} \cdot 1.2^8$$

Now, let's simplify that right hand side, it's numeric so we can just calculate it but I'll try and keep it as "easy" math. I assume the reader knows one of the following, though they're not too hard to quickly do by hand: $2^8 = 256$ or $2^{10} = 1024$ (which can be quickly derived from $2^8$)

$$\Rightarrow 1.2^8 = 10^{-8} \cdot 12^8$$ $$\Rightarrow 12^8 = 4^8 \cdot 3^8 = 2^{16} \cdot 3^8$$ $$\Rightarrow 3^8 = 9^4 = 81^2 = \left(80 + 1\right)^2 = 80^2 + 2\cdot 80 + 1 = 6400 + 160 + 1 = 6561$$ $$\Rightarrow 2^{16} = 2^{6} \cdot 2^{10} = 4 \cdot 4 \cdot 4 \cdot 1024 = 4 \cdot 4 \cdot 4096 = 4 \cdot 4(4100 - 4) = 4 \cdot 16384$$ $$\Rightarrow 1.2^8 = 10^{-8} \cdot 4 \cdot 10^{4} \cdot 1.6384 \cdot 10^4 \cdot 6.561 $$ $$\Rightarrow 4 \cdot 6.561 = 26.244 = 10 \cdot 2.6244$$ Meaning we can now simplify to $$\frac{\ln(2)}{\ln(5)} \stackrel{?}{>} 10^{-1} \cdot 1.2^8 = 10^{-1} \cdot 1.6384 \cdot 2.6244$$ Unfortunately, the numbers are too close to each other for us to use simple math or approximations anymore... Let's first get those logs, via the given for $\log_4(5) \approx 1.161$, noting that across multiplication we need to round up (it could be anywhere between 1.1610 and 1.1619) $$ \frac{\ln(2)}{\ln(5)} = \frac{1}{2 \cdot \frac{\ln(5)}{\ln(4)}} = \frac{1}{2 \cdot 1.161} = \frac{1}{2.323}$$ Inverting the expression we have $$ 2.323 \stackrel{?}{<} \frac{10}{1.6384 \cdot 2.6244}$$ As nice as it would be to just approximate to 1.7 times 2.7 or 1.64 times 2.63, we need to multiply up to 1.639 times 2.625 at least for precision. Note, it's rounded up to guarantee the result is smaller than the true value (since it's a fraction). $$\Rightarrow 1.639 \cdot 2.625 = 10^{-6} \cdot (1640 - 1) \cdot 2625$$ $$\Rightarrow 1640 \cdot 2625 = 10 \cdot 41 \cdot 4 \cdot 2625 = 10 \cdot 41 \cdot 10500 $$ $$ \Rightarrow 10^3 \cdot (40 + 1) \cdot 105 = 10^3 \cdot (4200 + 105) = 10^3 \cdot 4305 $$ $$ \Rightarrow 10^{-6} \cdot (10^3 \cdot 4305 - 2625) = 10^{-6} \cdot (4305000 - 2625) = 10^{-6} \cdot 4302375$$ Okay, just one more division to go. Trial division with that large number is pretty painful, but thankfully taking it as 4.3024 gives us enough precision, again rounding up. $$ 2.323 \stackrel{?}{<} \frac{100000}{43024} $$ To illustrate long division I will just list multiplications and remainders $$ \frac{100000}{43024} = \frac{86048 + 13952}{43024} = 2 + 10^{-1} \frac{129072 + 10448}{43024} = 2 + 0.3 + 10^{-2} \frac{86048 + 18432}{4303}$$ Our last remainder is above $4 \cdot 43024 = 172096$ so we have $$ \frac{100000}{43024} \geq 2 + 0.3 + 0.02 + 0.004 = 2.324 $$ Which means we have $ 2.323 \stackrel{?}{<} 2.324 $, which is true. Therefore, our original statement is true and $4^{5^9} > 5^{6^8}$.

Answer 8

Let $f=x^u$. If we take the derivative with respect to $x$, we get $df =ux^{u-1}dx$, which we can write as $$df =\frac u x f dx. \quad \cdots (1)$$ If we take the derivative with respect to $u$, we get $df=(\ln u) f du \cdots (2)$.

Now suppose we have $u=y^z$. We then get $du = \frac z y u dy$ and $du = (\ln z ) u dz$. If we plug those into $(2)$, we get (respectively) $$df = (\ln u) \frac z y uf dy \cdots (3)$$ and $$df = (\ln u)(\ln z ) ufdz \quad \cdots(4)$$

In $(1)$, $(3)$ and $(4)$ we can first disregard the common factor $uf$, and get $$\frac 1 x dx \quad \text{ vs } \quad (\ln u) \frac z y dy \quad \text{ vs } \quad (\ln u)(\ln z)dz.$$ The first one, $\frac 1 x dx$, is clearly the smallest of these, so we can then move on to compare the last two. These two have the common factor $(\ln u)$, and removing that leaves us with $$\frac z y dy\quad \text{ vs } \quad \ln z dz$$

If we plug in $x=4, y=5, z=9$, we get $\frac z y =1.8$ and $\ln z=2.19$. So $\ln z > \frac z y$.

This hold all along the way going to $x=5, y=6, z=8$. $z$ is decreasing and $y$ is increasing, so $\frac z y$ is decreasing. $\ln z$ monotonically decreases as $z$ decreases, so its minimum is at $z=8$. $\ln 8 =3\ln3 = 2.079$. The minimum value of $\ln z$ is greater than the maximum value of $\frac z y$, thus $\ln z > \frac z y$.

Going from $x=4, y=5, z=9$ to $x=5, y=6, z=8$, we have that $dx$ and $dy$ are positive, and $dz$ is negative, so $df$ is dominated by a decrease, and hence $f$ is decreasing.