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

1. ## 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?

2. ## 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

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

3. ## 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. )