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

why not

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

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

Printable View

- Feb 7th 2011, 05:19 PMalyosha2Negative Exponents / Algebraic Fractions / Reciprocal
$\displaystyle 3^4*2^{-8} = \frac{3^4}{2^8}$

why not

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

as $\displaystyle \frac{1}{2^8}$ is the reciprocal of $\displaystyle \frac{2^{-8}}{1}$ - Feb 7th 2011, 05:23 PMProve It
$\displaystyle \displaystyle 3^4 \times 2^{-8} = 3^4 \times \frac{1}{2^8} = \frac{3^4}{2^8}$.

- Feb 7th 2011, 05:23 PMdwsmith
- Feb 7th 2011, 11:16 PMalyosha2
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? - Feb 7th 2011, 11:33 PMalyosha2
Wait... $\displaystyle \frac{1}{2^8} = $ isn't the reciprocal of $\displaystyle 2^{-8}$, it's equivalent to it.

I see now.

Thanks.