1. ## exponents

find x so that 2^2^3^2^2 = 4^4^x

2. Originally Posted by sri340
find x so that 2^2^3^2^2 = 4^4^x
hi

$2^{2\cdot 3\cdot 2\cdot 2}=2^{2\cdot 2\cdot 2\cdot 2\cdot x}$

comparing , x=3

3. Hello, sri340!

With no parentheses, I assume those are "stacked" exponents.

$\text{Solve for }x:\;\;4^{4^x} \;=\;2^{2^{3^{2^2}}}$

$\text{The left side is: }\;4^{4^x} \;=\;4^{(2^2)^x} \;=\;4^{(2^{2x})} \;=\;(2^2)^{(2^{2x})} \;=\;2^{2\cdot2^{2x}} \;=\;2^{2^{2x+1}}$ .[1]

$\text{The right side is }\;2^{2^{3^{2^2}}} \;=\;2^{2^{3^4}} \;=\;2^{2^{81}}$ .[2]

$\text{Equate {\color{blue}[1]} and {\color{blue}[2]}: }\;2^{2^{2x+1}} \;=\;2^{2^{81}}$

$\text{Take logs (base 2): }\;\log_2\left(2^{2^{2x+1}}\right) \;=\;\log_2\left(2^{2^{81}}\right) \quad\Rightarrow\quad 2^{2x+1}\cdot\underbrace{\log_2(2)}_{\text{This is 1}} \;=\;2^{81}\cdot\underbrace{\log_2(2) }_{\text{This is 1}}$

$\text{We have: }\;2^{2x+1} \;=\;2^{81}$

$\text{Equate exponents: }\;2x + 1 \:=\:81 \quad\Rightarrow\quad 2x \:=\:80 \quad\Rightarrow\quad \boxed{x \:=\:40}$