How Can This Equation Simplification Be Achieved?

[lazypast]

$$w=\frac{k}{1-n}(v^{1-n}_{2}-v^{1-n}_{1})$$

let $$k=pv^n$$

$$\frac {p_{2}v^{n}_{2}v^{1-n}_{2} - p_{1}v^{n}_{1}v^{1-n}_{1}}{1-n}$$

how can this become

$$w=\frac {p_{1}v_{1} - p_{2}v_{2}}{n-1}$$


thanks

 

[Evalesco]

Hi lazypast,
Im not too sure how you got from the top equation to the middle equation but I'll assume it's all good.
You just need to use the indicie law $$a^{b}a^{c} = a^{(b+c)}$$
And then multiply through but a suitable form of 1, I reccomend $$\frac{-1}{-1}$$

 

[tacman]

To simplify the given equation, we can first substitute k with its equivalent value of pv^n. Then we can distribute the exponent of 1-n to both terms in the numerator, giving us p_{2}v^{n}_{2}v^{1-n}_{2} - p_{1}v^{n}_{1}v^{1-n}_{1}. Next, we can factor out the common terms of v^{n}_{2} and v^{n}_{1}, leaving us with v^{n}_{2}(p_{2}v^{1-n}_{2}) - v^{n}_{1}(p_{1}v^{1-n}_{1}). Finally, we can use the distributive property in reverse and factor out the common term of (n-1) from the remaining terms, giving us w=\frac{(p_{2}v^{1-n}_{2}-p_{1}v^{1-n}_{1})(v^{n}_{2}-v^{n}_{1})}{n-1}. This can then be simplified further to w=\frac{p_{1}v_{1} - p_{2}v_{2}}{n-1}, as requested.