What algebra rule is used here to give the exponent 2 in step 2?

[hellothere2]

First it starts as

r= p* (50K^-.5 100^.5)

then K=[(50p100^.5/r]^2

So how does the power of 2 get there in the second part when moving K to the other side?

 

[Ackbach]

You have to go step-by-step in solving for $K$:
\begin{align*}
r&= p (50K^{-0.5} 100^{0.5}) \\
\frac{r}{50p}&=K^{-0.5} 10 \\
\frac{r}{500p}&=\frac{1}{K^{0.5}} \\
\frac{500p}{r}&=K^{0.5} \\
\left(\frac{500p}{r}\right)^{\!2}&=K.
\end{align*}
So to answer your question, the $2$'s come about when you square both sides as the last step.

 

[soroban]

Hello, hellothere!

[tex]r \:=\: p (50K^{-0.5}10^{0.5})[/tex]

then: [tex] K\:=\:\left(\frac{50p100^{\frac{1}{2}}}{r}\right)^2[/tex]

So how does the power of 2 get there?

We have: .[tex]r \;=\;p(50K^{-\frac{1}{2}}100^{\frac{1}{2}}) [/tex]

Multiply by [tex]\frac{K^{\frac{1}{2}}}{r}\!:\;\;K^{\frac{1}{2}} \;=\;\frac{50p100^{\frac{1}{2}}}{r}[/tex]

Square both sides: [tex]\;K \;=\;\left(\frac{50p100^{\frac{1}{2}}}{r}\right)^2[/tex]