Which giant number is greater?

I would not expect a correct answer to have the word "clearly" in it. How can I know that $3^{61} > 2^{64}$ without calculations?

Nice solution and question.

I completely misread this. I thought it was $[3^5]^3$ . Oops...

I was trying another way which looked good at first, but I don't know how to go on

$6=3^{\log_3{6}}$

So we have

$3^{125}$ and $3^{ \log_3(6) 64}$

which brings us to

$125 > \log_3(6) 64$

Which is true if

$\log_3{6}<\frac{125}{64}$

$6<3^{\frac{125}{64}}$

but $3^{\frac{125}{64}}>3^{1+\frac{1}{2}+\frac{1}{4}}=3 * 3^{1/2} * 3^{1/4}$

is there a "polite" way to estimate those numbers without brutal calculation?

why exactly 3^{125} and 6^{64} ?

${6^{{2^6}}}\quad = \quad {2^{{2^6}}} \times {3^{{2^6}}}\quad < \quad {3^{{2^6} + 1}}\quad = \quad {3^{{4^3} + {1^3}}}\quad < \quad {3^{{5^3}}}$

I interpreted the expressions in the problem statement like so: [(a^b)^c] becomes [a^(b*c)]. Now I know what the colors of "b" and "c" terms mean.

2^64=2^9.2^55=512.2^55<729.3^55=3^61

Here's how I see the problem. When you have an exponent to another exponent, which is the case with this problem you use the Power Rule for Exponents. The power rule says to multiply the exponents and keep the base:

$(x^m)^{n} = x^{m^n} = x^{m \cdot n}$ . This same applies for this problem. Using the power rule,

$3^{5^3} = (3^5)^3 = 3^{15}$ and $6^{2^6} = (6^2)^6 = 6^{12}$ . You can rewrite $6^{12}$ as $(2 \cdot 3) ^ {12} = 2^{12} \cdot 3^{12}$ using the "product to a power rule". Now we can compare the two number by rewriting:

$3^{15} = 3^3 \cdot 3^{12}$ _ _ _ _ $2^{12} \cdot 3^{12} = 6^{12}$ . I obtained the left hand side by using the product rule for exponents in reverse. Now I can divide both sides of this inequality by $3^{12}$ , assuming that both sides are not same, but even if they are I can still do this. Now I can compare $3^3 = 27$ and $2^{12} = 4096$ . Clearly $27 < 4096$ , therefore the other number, $6^{2^6}$ is larger.