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

  • Thread starter Thread starter hellothere2
  • Start date Start date
  • Tags Tags
    Algebra Exponent
AI Thread Summary
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.
hellothere2
Messages
1
Reaction score
0
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?
 
Mathematics news on Phys.org
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
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Replies
6
Views
2K
Replies
15
Views
3K
3
Replies
105
Views
6K
Replies
5
Views
2K
Replies
7
Views
3K
Back
Top