# Thread: Negative Exponents / Algebraic Fractions / Reciprocal

1. ## Negative Exponents / Algebraic Fractions / Reciprocal

$3^4*2^{-8} = \frac{3^4}{2^8}$

why not

$\frac{3^4}{\frac{1}{2^8}}$

as $\frac{1}{2^8}$ is the reciprocal of $\frac{2^{-8}}{1}$

2. $\displaystyle 3^4 \times 2^{-8} = 3^4 \times \frac{1}{2^8} = \frac{3^4}{2^8}$.

3. Originally Posted by alyosha2
$3^4*2^{-8} = \frac{3^4}{2^8}$

why not

$\frac{3^4}{\frac{1}{2^8}}$

as $\frac{1}{2^8}$ is the reciprocal of $\frac{2^{-8}}{1}$
$\displaystyle 3^4\cdot\frac{1}{2^8}=\frac{3^4}{1}\cdot\frac{1}{2 ^8}=\frac{3^4\cdot 1}{1\cdot 2^8}=\frac{3^4}{2^8}$

4. Originally Posted by Prove It
$\displaystyle 3^4 \times 2^{-8} = 3^4 \times \frac{1}{2^8} = \frac{3^4}{2^8}$.
So multiplying by a given number gives the same result as multiplying it by it's reciprocal!??

I thought the idea was to invert the operation? Meaning dividing by a given number is the same as multiplying by it's reciprocal and dividing by a given number is the same as multiplying by it's reciprocal?

If that's the case, how come the operation on the number here isn't inverted when we subsitutue it for it's reciprocal?

5. Wait... $\frac{1}{2^8} =$ isn't the reciprocal of $2^{-8}$, it's equivalent to it.

I see now.

Thanks.