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 $$\ln\frac{\ln5}{\ln4} < \ln 1.1612 < 0.1612 - \frac{0.1612^2}{2} + \frac{0.1612^3}{3} < 0.1497$$ (it is an upper bound since we have just added a positive term, the cubed term, to an alternating series). 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

(This may be construed as bending the rules, but I assume that divisions can be carried out to four decimals by hand.)
For starters,

$4^5 > 5^{1.2^8}$, so the left side is bigger, and we 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
$2.) \ln\left(\frac{n+1}{n}\right) < \frac{2}{2n+1}+\frac{1}{6n(n+1)(2n+1)}$

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!