How do you evaluate 8^0 x 2^-3?

Apr 7, 2016

${8}^{0} x {2}^{- 3} = \frac{x}{2} ^ \left(3\right)$

Explanation:

$\textcolor{b l u e}{\text{Explaining about } {8}^{0}}$

Note that ${8}^{0} = 1$

Note that $2 = {2}^{1}$

Suppose we had $\frac{4}{2} = 2$. But $4 = {2}^{2}$ so we have ${2}^{2} / {2}^{1}$

It is allowed that you do this: ${2}^{2} / {2}^{1} = {2}^{2 - 1} = {2}^{1} = 2$

Suppose we had $\frac{2}{2} = 1$ this is the same as ${2}^{1} / {2}^{1} = {2}^{1 - 1} = {2}^{0} = 1$

So anything raise to the power of 0 is one. However, there is a lot of debate about ${0}^{0}$ so stay clear of that one!

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Given:$\text{ } {8}^{0} x {2}^{- 3}$

Known:${8}^{0} = 1$

so ${8}^{0} x {2}^{- 3} = x {2}^{- 3}$

But ${2}^{- 3}$ is the same as $\frac{1}{2} ^ 3$ giving:

${8}^{0} x {2}^{- 3} = \frac{x}{2} ^ \left(3\right)$