# Rules of exponents $(\frac{u}{v})^{a+b-1}=u^{a+b-1}\cdot v^{-(a+b-1)}$

I'm taking undergraduate statistics this semester. We have a problem concerning the beta distribution. In which there are some exponent laws in use that I'm having a hard time understanding. In my professor's solutions he makes this step.

\begin{equation*} \left(\frac{u}{v}\right)^{a+b-1}=u^{a+b-1}\cdot v^{-(a+b-1)} \end{equation*}

and this one

\begin{equation*} \left(1-\frac{u}{v}\right)^{c-1}=v^{-(c-1)}\cdot(v-u)^{c-1} \end{equation*}

Can anyone help me understand how these equate?

• Note that the originally asked question had the first equality not true. It should read $(u/v)^{a+b-1}=u^{a+b-1}*v^{-(a+b-1)}$ NOT $(u/v)^{a+b-1}=u^{a+b+1}*v^{-(a+b-1)}$ – Yeah.. Feb 13 '17 at 16:03

Well note that $$\left( \frac{x}{y} \right)^n=\frac{x^n}{y^n}$$ And $y^{-n}=\frac{1}{y^n}$. So $$\left( \frac{x}{y} \right)^n=x^{n}y^{-n}$$ The ones you mention follow. For example,

\begin{equation*} \left(1-\frac{u}{v}\right)^{c-1}=\left(\frac{v-u}{v}\right)^{c-1}=v^{-(c-1)}\cdot(v-u)^{c-1} \end{equation*}

$$\left(\frac{u}{v}\right)^{a+b-1} = u^{a+b-1}\left(\frac{1}{v}\right)^{a+b-1} = u^{a+b-1}v^{-\left({a+b-1}\right)}$$

In other words, exponents distribute over multiplication, and one divided by a number raised to a power is the same as that number raised to the negative power.

For the second example,

$$1-\frac{u}{v} = \frac{v}{v} - \frac{u}{v} = \frac{v-u}{v}$$

Then, the same rules that I mentioned above apply.

$$(u/v)^{a+b-1}=u^{a+b-1}*v^{-(a+b-1)}$$ $$(1-u/v)^{c-1}=v^{-(c-1)}*v^{c-1}*(1-u/v)^{c-1}=v^{c-1}*(v*(1-u/v))^{c-1}=v^{c-1}*(v-u)^{c-1}$$