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!
 A: Notice that $a^{-n}=\frac{1}{a^n}$ and $a^b \cdot a^c = a^{b+c}$.
$$\frac{1}{2^{-3}}\cdot 2^{-10}=\frac{1}{2^{-3}}\cdot \frac1{2^{10}}=\frac1{2^{10-3}}$$
A: There is usually more than one way to simplify an expression.
I would probably have chosen to multiply numerator and denominator by $2^3$ in order to deal with exponents only and not with fractions.
Another approach would be to multiply numerator and denominator by $2^{10},$
which would cancel the $2^{-10}$ on the right of the fraction.
But a slightly different approach was taken by whoever did it the way you saw. Apparently they decided to divide by $2^{-10},$ which has the same effect as multiplying by $2^{10}$, but the direct interpretation of $2^a$ divided by $2^b$ is $2^{a-b},$ and in this example $a=-3$ and $b=-10,$
so $a-b$ is literally $-3-(-10).$
But for goodness sake, why didn’t they simplify away the division by $1$ before tackling the rest of it?
A: $b^nb^m = b^{n+m}$ 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}= $
$(2\cdot 3)^6 \cdot 2^{-7}=$
$2^6\cdot 3^6 \cdot 2^{-7}=$
$2^{-1}\cdot 3^6=$
$\frac {3^6}2$.
=====
Or we could simply do:
$\frac {\frac {6^6}1}{2^{-3}} \cdot 2^{-10}$
$\frac {\frac {6^6}1}{\frac 1{2^3}}\cdot \frac 1{2^{10}}=$
$\frac {\frac {6^6}1}{\frac 1{2^3}2^{10}}=$
$\frac {\frac {6^6}1}{2^7}$.
A: \begin{align}
\frac{\;\;\dfrac{6^6}{1}\;\;}{2^{-3}}\cdot2^{-10}
&=\frac{6^6}{2^{-3}}\cdot 2^{-10}\qquad&\text{(simplify the denominator)}\\
&=6^6\cdot 2^3\cdot 2^{-10}\qquad &(\frac{1}{2^{-3}}=2^3)\\
&=6^6\cdot 2^{3+(-10)}\qquad &(2^a2^b=2^{a+b})\\
&=6^6\cdot 2^{-7}\qquad &\text{(simplify)}\\
&=\frac{6^6}{2^7}\qquad &(2^{-7}=\frac{1}{2^7})
\end{align}
A: Making all exponents positive,
$$\frac{\dfrac{6^6}{1}}{2^{-3}}2^{-10}=\frac{6^6\cdot 2^3}{1\cdot2^{10}}=\frac{3^6\cdot2^6\cdot2^3}{2^{10}}=\frac{729}2.$$
A: I think it's a problem of confusing notation. If you read the expression as (which is what I first assumed is written)
$$
6^6\div\frac{1}{2^{-3}}\cdot 2^{-10}=\frac{729}{128}.
$$
There is no simplification like the one you wrote. However, if you read it as
$6^6\cdot\frac{1}{2^{-3}}\cdot 2^{-10}=\frac{729}{2}=\frac{6^6}{2^7}$ you get the right answer. I'm not really sure why the question bothered to divide $6^6$ by 1.
A: First off, the way in which the problem is typeset is almost intended to cause confusion.  Adding some parentheses and not using \dfrac everywhere should make it easier to understand the expression:
$$\frac{\left(\frac{6^6}{1}\right)}{2^{-3}} \cdot 2^{-10}. $$
Division is the inverse operation of multiplication and so, rather than thinking of the problem as a division problem, it might be easier to think of it as the multiplication of several fractions.  Specifically,
$$\frac{\left(\frac{6^6}{1}\right)}{2^{-3}} \cdot 2^{-10}
= 6^6 \cdot \frac{1}{2^{-3}} \cdot 2^{-10}. $$
One of the fundamental properties of exponents is that (under appropriate hypotheses) $a^{-n} = \frac{1}{a^n}$, and so we can further simplify this expression to
$$ 6^6 \cdot \frac{1}{2^{-3}} \cdot 2^{-10}
= 6^6 \cdot 2^{3} \cdot \frac{1}{2^{10}}
= 6^6 \cdot \frac{2^{3}}{2^{10}}. $$
Finally, another property of exponents is that
$$ \frac{a^m}{a^n} = a^{m-n}. $$
Applying this to the last result gives
$$ 6^6 \cdot \frac{2^{3}}{2^{10}}
= 6^6 \cdot 2^{3-10}
= 6^6 \cdot 2^{-7}
\overset{\text{or}}{=} \frac{6^6}{2^7}. $$
Personally, I would also probably use the fact that $(ab)^n = a^n b^n$ to rewrite $6^6 = 2^6\cdot 3^6$, and conclude by writing
$$ 6^6 \cdot 2^{-7} = 2^6\cdot 3^6 \cdot 2^{-7}
= 2^{6-7} \cdot 3^6
= 2^{-1} \cdot 3^6
\overset{\text{or}}{=} \frac{3^6}{2}, $$
but perhaps this is not part of the intended exercise.
