Which is larger? 4^(5^9) or 5^(6^8)

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.

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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

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 3

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 enough to prove

$$1.024^{14}\cdot 8 \stackrel{?}{>} 10$$ 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 4

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 $2n+1$ 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 5

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 6

Based on $5$, $4^{1161} \approx 5^{1000}$.

We need to compare $4^{ln^{1.6094 \times 9}} = 4^{ln^{14.4846}}$ and $5^{ln^{(1.0986+ 0.6931)\times 8}} = 5^{ln^{1.7917\times 8}} = 5^{ln^{14.3336}}$, therefore:

$4^{ln^{14.4846}}$ to $5^{ln^{14.3336}}$
$4^{ln^{14.4846}\times1000}$ to $4^{ln^{14.3336}\times1161}$
$ln^{14.4846}\times1000$ to $ln^{14.3336}\times1161$
$ln^{0.1510}\times1000$ to $1161$
$ln^{0.1510}\times1000=ln^{0.1510}\times ln^{(0.6931+1.6094)\times3}$ to $43\times ln^{1.0986\times3}$
$ln^{7.0585}$ to $43\times ln^{3.2958}$
$ln^{3.7627}$ to $43$

We can work out that $ln(216)=5.3751$ and $ln(225)=5.416$. These logarithms grow increasingly slowly, so:

$ln(215)<5.3751-(5.416-5.3751)/9 => ln(215)<5.3706$
$ln(43)<5.3706-ln(5) => ln(43)<3.7612$

So, the left side is slightly larger.