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Thread: Is b^2^3=(b^2^2)^2?

  1. #1
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    Is b^2^3=(b^2^2)^2?

    Stupid question of the week. My book says "notice each number in
    $\displaystyle b,b^2,{b^2}^2,{b^2}^3,{b^2}^4\ldots$
    is the square of the preceding one"

    In other words
    $\displaystyle {b^2}^3={({b^2}^2)}^2$.

    Fortunately,
    $\displaystyle {({b^2}^2)}^2=({b^2}^2)({b^2}^2)={b^{2^2+2^2}={b^{ 2 \cdot2^2}={b^2}^3$. \checkmark (or \checkbox, or...whatEVER).

    But also
    $\displaystyle {({b^2}^2)}^2={(b^4)}^2=(b^4)(b^4)=b^8$.
    I mean, two squared is usually four, right?

    Isn't $\displaystyle {b^2}^3={({b^2}^2)}^2$? (Actually, I know the answer: Yes.)

    So what's wrong with my "But also..." line?
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  2. #2
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    Re: Is b^2^3=(b^2^2)^2?

    It's ok.

    The book is just showing the pattern

    $\displaystyle b^{2^{(n+1)}}=\left(b^{2^n}\right)^2$

    since

    $\displaystyle 2^{n+1}=(2)2^n$
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  3. #3
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    Re: Is b^2^3=(b^2^2)^2?

    Oh, wait a minute (brain fart, or somethin' like that ).

    $\displaystyle b^8={b^2}^3$

    Somehow, on paper, I was getting b^6. Then I typed it in here, and mindlessly calculated the correct value.

    (I knew this was going to be an embarrassing question. Oh well. )
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