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I have this equation given $1=\alpha^{(CW-1)} N$ and it solution has been given as $\alpha=N^{ \frac{-1}{CW-1}} $.
I want to obtain same expression for $\alpha$ but i am unable to get it. I tried to solve it by taking natural log (ln) on both sides and i get something like this $\alpha=\frac{-N}{CW-1}$.

Can any body tell what mistake i have made and what steps i should do to get the correct solution.

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  • $\begingroup$ If $a^n = b$, then $a=b^{1/n}$, $n \ne 0$ $\endgroup$
    – John_dydx
    Apr 18, 2020 at 19:00
  • $\begingroup$ Perhaps you should have gotten $\ln \alpha = \frac{-\ln N}{CW-1}$? $\endgroup$
    – John Joy
    Apr 18, 2020 at 21:30
  • $\begingroup$ Is you queston : through which steps do we get the solution that is given? $\endgroup$
    – user655689
    Apr 18, 2020 at 22:52

1 Answer 1

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You can write:

$\begin{align*} N &= \alpha^{-(C W - 1)} \\ N^{-\frac{1}{C W - 1}} &= \alpha \end{align*}$

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  • $\begingroup$ Thanks what if i had $4=\alpha^{(CW-1)} N$ $\endgroup$
    – user3696623
    Apr 18, 2020 at 19:03

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