Is it equivalent to raising every term by 4?

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

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

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

So the answer is 64.

Let's answer this question more generally for $\underbrace{n^n\times n^n\times \cdots \times n^n}_{N\ \text{times}} = n^{n^n}.$ Taking the $n$ -logarithm on both sides, the multiplication becomes addition: $\underbrace{n+ n + \cdots + n}_{N\ \text{times}} = n^n.$ Thus $Nn = n^n\ \ \ \ \therefore\ \ \ \ \ N = \frac{n^n}n = n^{n-1}.$ In this case, $N = 4^{4-1} = 4^3 = \boxed{64}$ .

$4^{4^{4}}=4^{4 \times 4^{3}}=\left(4^{4}\right)^{4^{3}}$

And so the factor $4^{4}$ is repeated $4^{3}=4 \times 4 \times 4 = \boxed{64}$ times.

Say that there are $x$ terms on the left side of the equation.

Using the rules of exponents, we know $a^m \times a^n=a^{m+n}$

Getting rid of the base, we can set up an equation: $4x=4^4=256$

If we divide $x$ from both sides, we get the solution $x=64$ .

Note: the left side of the first equation is saying that there are $x$ 4's that are being added together.

$4^{(4^{4}})=4^{4\cdot 4\cdot 4\cdot 4}=(4^{4})^{4^{3}}=(4^{4})^{\boxed{64}}$

Since $4^4$ is $256$ , by the product rule $\frac{256}{4}$ would be the answer $64$ .

$4^{4} = 64$

$\overbrace {4 \times 4 \times \ldots 4 \times 4}^{64}$

So, as always $\boxed {64}$

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

It's actually quite simple once you break it down : 4(2) = 4x4 =16 4(3) = 4x4x4 = 64. n =64

We just went over a concept just like this in my algebra 2/ trig class (I'm still in high school). If I'm being honest I don't really know what I did but I got the answer right so I guess there's that going for me.

Using the Natural exponent definition from the introduction, the left side becomes

(4^4)^n

= 4^(4*n) = 4^(4^4)

So,

4*n = 4^4

n = 4^3 = 64