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

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The discussion focuses on how the exponent of 2 is derived in the equation for K. Initially, the equation is expressed as r = p(50K^(-0.5)100^(0.5)). To isolate K, the steps involve manipulating the equation by multiplying and rearranging terms. The power of 2 appears when both sides of the equation are squared to solve for K. Ultimately, the exponent of 2 is introduced as a result of squaring the equation after isolating K.
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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?
 
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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.
 
Hello, hellothere!

r \:=\: p (50K^{-0.5}10^{0.5})

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

So how does the power of 2 get there?
We have: .r \;=\;p(50K^{-\frac{1}{2}}100^{\frac{1}{2}})

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

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