Return to Answer

Starting with the claim

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

We have

$$ \iff 5^9\ln4 > 6^8\ln5 $$ $$ \iff \frac{5^9}{6^8} > \frac{\ln5}{\ln4} = 1.161 $$ $$ \iff 6\times(5/6)^9 > \frac{\ln5}{\ln4} $$ $$ \iff \ln(6)+9\ln(5/6) > \ln\frac{\ln5}{\ln4} $$ Using $\ln x = - \ln(1/x)$, $$ \iff \frac{\ln6 - \ln\frac{\ln5}{\ln4}}{\ln(6/5)} > 9 $$

Using the hints we have

$$ \frac{\ln5}{\ln4} = 1.161 $$ $$ \ln(4) = 2\ln(2) = 2\times0.6931 $$ $$ \ln(6) = \ln(2)+\ln(3) = 0.6931+1.0986 = 1.7917 $$ $$ \ln(1.161) = \ln(1+0.161) < 0.161 - \frac{0.161^2}{2} + \frac{0.161^3}{3} = 0.1494 $$ $$ \ln(6) = 1.7917 $$ $$ \ln(6/5) = \ln(6)-\ln(5) = 1.7917 - 1.6094 = 0.1823 $$

we get that the claim

$$ \iff \frac{\ln6 - \ln\frac{\ln5}{\ln4}}{\ln(6/5)} \approx \frac{1.7917 - 0.1494}{0.1823} = 9 + \frac{16}{1823}> 9$$

which is true

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

Starting with the claim

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

We have

$$ \iff 5^9\ln4 > 6^8\ln5 $$ $$ \iff \frac{5^9}{6^8} > \frac{\ln5}{\ln4} = 1.161 $$ $$ \iff 6\times(5/6)^9 > \frac{\ln5}{\ln4} $$ $$ \iff \ln(6)+9\ln(5/6) > \ln\frac{\ln5}{\ln4} $$ Using $\ln x = - \ln(1/x)$, $$ \iff \frac{\ln6 - \ln\frac{\ln5}{\ln4}}{\ln(6/5)} > 9 $$

Using the hints we have

$$ \frac{\ln5}{\ln4} = 1.161 $$ $$ \ln(4) = 2\ln(2) = 2\times0.6931 $$ $$ \ln(6) = \ln(2)+\ln(3) = 0.6931+1.0986 = 1.7917 $$ $$ \ln(1.161) = \ln(1+0.161) < 0.161 - \frac{0.161^2}{2} + \frac{0.161^3}{3} = 0.1494 $$ $$ \ln(6) = 1.7917 $$ $$ \ln(6/5) = \ln(6)-\ln(5) = 1.7917 - 1.6094 = 0.1823 $$

we get that the claim

$$ \iff \frac{1.7917 - 0.1494}{0.1823} = 9 + \frac{16}{1823}> 9$$

which is true

Starting with the claim

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

We have

$$ \iff 5^9\ln4 > 6^8\ln5 $$ $$ \iff \frac{5^9}{6^8} > \frac{\ln5}{\ln4} = 1.161 $$ $$ \iff 6\times(5/6)^9 > \frac{\ln5}{\ln4} $$ $$ \iff \ln(6)+9\ln(5/6) > \ln\frac{\ln5}{\ln4} $$ Using $\ln x = - \ln(1/x)$, $$ \iff \frac{\ln6 - \ln\frac{\ln5}{\ln4}}{\ln(6/5)} > 9 $$

Using the hints we have

$$ \frac{\ln5}{\ln4} = 1.161 $$ $$ \ln(4) = 2\ln(2) = 2\times0.6931 $$ $$ \ln(6) = \ln(2)+\ln(3) = 0.6931+1.0986 = 1.7917 $$ $$ \ln(1.161) = \ln(1+0.161) < 0.161 - \frac{0.161^2}{2} + \frac{0.161^3}{3} = 0.1494 $$ $$ \ln(6) = 1.7917 $$ $$ \ln(6/5) = \ln(6)-\ln(5) = 1.7917 - 1.6094 = 0.1823 $$

we get that the claim

$$ \iff \frac{\ln6 - \ln\frac{\ln5}{\ln4}}{\ln(6/5)} \approx \frac{1.7917 - 0.1494}{0.1823} = 9 + \frac{16}{1823}> 9$$

which is true

Starting with the claim

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

We have

$$ \iff 5^9\ln4 > 6^8\ln5 $$ $$ \iff \frac{5^9}{6^8} > \frac{\ln5}{\ln4} = 1.161 $$ $$ \iff 6\times(5/6)^9 > \frac{\ln5}{\ln4} $$ $$ \iff \ln(6)+9\ln(5/6) > \ln\frac{\ln5}{\ln4} $$ Using $\ln x = - \ln(1/x)$, $$ \iff \frac{\ln6 - \ln\frac{\ln5}{\ln4}}{\ln(6/5)} > 9 $$

Using the hints we have

$$ \frac{\ln5}{\ln4} = 1.161 $$ $$ \ln(4) = 2\ln(2) = 2\times0.6931 $$ $$ \ln(6) = \ln(2)+\ln(3) = 0.6931+1.0986 = 1.7917 $$ $$ \ln(1.161) = \ln(1+0.161) < 0.161 - 0.161^2/2 + 0.161^3/3 = 0.1494 $$ $$ \ln(6) = 1.7917 $$ $$ \ln(6/5) = \ln(6)-\ln(5) = 1.7917 - 1.6094 = 0.1823 $$

we get that the claim

$$ \iff (1.7917 - 0.1494)/0.1823 > 9$$ $$ \iff 9.0087 > 9$$

which is true

Starting with the claim

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

We have

$$ \iff 5^9\ln4 > 6^8\ln5 $$ $$ \iff \frac{5^9}{6^8} > \frac{\ln5}{\ln4} = 1.161 $$ $$ \iff 6\times(5/6)^9 > \frac{\ln5}{\ln4} $$ $$ \iff \ln(6)+9\ln(5/6) > \ln\frac{\ln5}{\ln4} $$ Using $\ln x = - \ln(1/x)$, $$ \iff \frac{\ln6 - \ln\frac{\ln5}{\ln4}}{\ln(6/5)} > 9 $$

Using the hints we have

$$ \frac{\ln5}{\ln4} = 1.161 $$ $$ \ln(4) = 2\ln(2) = 2\times0.6931 $$ $$ \ln(6) = \ln(2)+\ln(3) = 0.6931+1.0986 = 1.7917 $$ $$ \ln(1.161) = \ln(1+0.161) < 0.161 - \frac{0.161^2}{2} + \frac{0.161^3}{3} = 0.1494 $$ $$ \ln(6) = 1.7917 $$ $$ \ln(6/5) = \ln(6)-\ln(5) = 1.7917 - 1.6094 = 0.1823 $$

we get that the claim

$$ \iff \frac{1.7917 - 0.1494}{0.1823} = 9 + \frac{16}{1823}> 9$$

which is true