<s>[INST] How does this division work? $\frac{\;\frac{6^6}{1}\;}{2^{-3}}\cdot2^{-10}$ I came across this division and can't wrap my head around how it is solved:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^{-3}}\cdot2^{-10}$$
They subtract the exponent of $2^{-10}$ from the denominator's exponent $2^{-3}$:
$$2^{-3-(-10)}$$
Which gives us:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^7}$$
If anyone knows what is actually being done here I would appreciate it!
 [/INST] $b^nb^m = b^{n+m}$ and $b^{-n} = \frac 1b$ and $\frac {b^n}{b^m} = b^{n-m} = \frac 1{b^{m-n}}$.  That is all that is going on.
$\frac {BLAH}{\frac 1{2^{-3}}\cdot 2^{-10}}=$
$\frac {BLAH}{\frac 1{2^{-3}}\frac 1{2^{-(-10)}}}=\frac {BLAH}{\frac 1{2^{-3 - (-10)}}}$
Frankly seems like a convoluted way to make things as hard as possible and to make the negative signs as many and as confusing as possible.
I'd have just done:
$\frac{6^6}{\frac 1{2^{-3}}\cdot 2^{-10}}=$
$\frac {(2\cdot 3)^6}{2^3\cdot 2^{-10}}=$
$\frac {2^6\cdot 3^6}{2^{-7}}=$
$2^6\cdot 3^6 \cdot 2^7 = 2^{13}\cdot 3^6$.
.....
Oh.... I now see the expression was supposed to be $\frac {\frac {6^6}1}{2^{-3}}\cdot 2^{-10}$ and not $\frac{6^6}{\frac 1{2^{-3}}\cdot 2^{-10}}$.
That doesn't change me comments and answer.  But sheesh what kind of lunatic wrote this problem solely for the purpose of confusion?
We have $\frac {BLAH}{2^{-3}}\cdot 2^{-10} =$
$\frac {BLAH}{2^{-3}2^{-(-10)}} = \frac {BLAH}{2^{-3-(-10)}}$.
But I'd do:
$\frac {\frac {6^6}1}{2^{-3}}\cdot 2^{-10}=$
$\frac {6^6}{2^{-3}}\cdot 2^{-10}=$
$6^6\cdot 2^3 \cdot 2^{-10} =$
$6^6 \cdot 2^{-7}= $
$(2\cdot 3)^6 \cdot 2^{-7}=$
$2^6\cdot 3^6 \cdot 2^{-7}=$
$2^{-1}\cdot 3^6=$
$\frac {3^6}2$.
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Or we could simply do:
$\frac {\frac {6^6}1}{2^{-3}} \cdot 2^{-10}$
$\frac {\frac {6^6}1}{\frac 1{2^3}}\cdot \frac 1{2^{10}}=$
$\frac {\frac {6^6}1}{\frac 1{2^3}2^{10}}=$
$\frac {\frac {6^6}1}{2^7}$.
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