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Math Help - exponents

  1. #1
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    exponents

    find x so that 2^2^3^2^2 = 4^4^x
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  2. #2
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    Quote Originally Posted by sri340 View Post
    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
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  3. #3
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    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}

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