# How do you rewrite 1/(4b^3) using negative exponents?

Jan 2, 2016

${4}^{- 1} {b}^{- 3}$

#### Explanation:

Suppose we had $\frac{1}{x}$. This would have a particular value. If you turned it upside down to give $\frac{x}{1}$ you have changed in inherent value so you need to apply a conversion.

The conversion used in this example is ${x}^{- 1}$.

Although this looks deferent it has the same value as $\frac{1}{x}$

$\textcolor{g r e e n}{\text{Note that "x" is the same as } {x}^{1}}$
so $\frac{1}{x} ^ 1 = {x}^{- 1}$

What if we had $\frac{1}{{x}^{3}}$ then we would write: ${x}^{- 3}$
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Dealing with your question using the same logic.

Given : $\frac{1}{4 {b}^{3}}$

$\textcolor{b l u e}{\text{Assumption}}$

color(brown)(underline("Everything")" is meant to have negative exponents")

Initially write it as: $\frac{1}{4} \times \frac{1}{b} ^ 3$

Now write as: $\frac{1}{4} \times {b}^{- 3} / 1 = \frac{1 \times {b}^{- 3}}{4 \times 1}$

So now we have: ${b}^{- 3} / 4$

You could take this further by starting to 'play' with $\frac{1}{4}$

One potential format would be: ${4}^{- 1} {b}^{- 3}$

Another: another way of writing 4 is ${2}^{2}$

So we could write: $\textcolor{w h i t e}{. .} {4}^{- 1} {b}^{- 3} \textcolor{w h i t e}{.} \text{ as } \textcolor{w h i t e}{.} {2}^{- 2} {b}^{- 3}$

Which is equally true!