Is it equivalent to raising every term by 4?

Question

We know that $(\underbrace{4 \times 4 \times 4 \times 4}_{4 \text{ times}} = 4^4.$

But how many $(4^4)$ are there in the left side of the equality $\underbrace{4^4 \times 4^4 \times \cdots \times 4^4}_{\text{How many times?}} = {\large 4^{\left( 4^4 \right)}} ?$

Answer 1

Let the number of $4^4$ on the LHS be $n$. Then:

$\large \begin{align} \underbrace{4^4 \times 4^4 \times 4^4 \times \cdots \times 4^4}_{\text{Number of }4^4 = n} & = 4^{4^4} \\ 4^{\overbrace{4+4+4+\cdots+4}^{\text{Number of }4 = n}} & = 4^{4^4} \\ 4^{4n} & = 4^{4^4} \\ \implies 4n & = 4^4 \\ n & = 4^3 = \boxed{64} \end{align} $

This problem is similar to $ \large \sqrt{ 2^{ \sqrt{2} } } = \, ? $ in attempting to get the solver to use an invalid exponent rule. The specific rule is

$(a^b)^c = (a^c)^b$

not

$a^{(b^c)} = a^{(c^b)}$

So while $ (4^4)^4 $ has 4 copies of $ 4^4 ,$ $ 4^{(4^4)} $ does not.

It may be helpful to remember this applies to other non-commutative operations like subtraction. For example, $ 10 - (9 - 1) = 2 $ but $10 - (1 - 9) = 18 .$

Answer 2

$$4^4 \times ... \times 4^4 = 4^{(4^{4})} \Longrightarrow (4^{4})^{n} = 4^{(4^{4})} \Longrightarrow 4n = 4^{4} \Longrightarrow n = 64$$

Answer 3

Since $4^4=4\cdot 4^3$, we have $$ 4^{4^4}=4^{4\cdot4^3}=(4^4)^{4^3} $$

Answer 4

$4^4×4^4×4^4.........=4^{4^4}.$

Rewrite:

$4×4×4.........= 4^{4^4}, $

You have $4^4$ factors $4$ on the LHS of the above equation.

Group them into $4$:

$(4×4×4×4)(4×4×4×4)............= 4^{4^4};$

How many parentheses $ (4×4×4×4)= 4^4$ are there?

Answer: $4^4/4=4^3= 64.$

Answer 5

$ 4^{4^{4}} = 4^{256} = (4^{4})^{64}$

So the answer is $64$.