General algebra doubt

Let $$\left(\frac{\cos\alpha}{\sin\alpha}\right) = \left(\frac{b}{a}\right)^n \left(\frac{x_0}{y_0}\right)^\frac{1}{n - 1}$$

Now suppose if I multiply by power $(n - 1)$ to both sides then the equation will look something like this :

$$\left(\frac{\cos\alpha}{\sin\alpha}\right)^\left(n - 1\right) = \left(\frac{b}{a}\right)^n \left(\frac{x_0}{y_0}\right)$$

Have I committed some mistake here? I'm asking this because I'm confused. Also I'm not getting correct answer of my question. I think I have did all the steps correctly. I just have doubt in this one.

The doubt is whether multiplying by power $(n - 1)$ to both sides then l.h.s is easy but how will the $\left(\frac{b}{a}\right)^n$ term on r.h.s will be effected. Kindly point out my mistake.

• you forgot to raise $(b/a)^n$ to the power as well – Zach Effman Mar 17 '16 at 16:38
• @ZachEffman where? – Hisenberg Mar 17 '16 at 16:41
• In the only step you show here – Zach Effman Mar 17 '16 at 16:42
• Also $(\frac{x_0}{y_0})^{(\frac{1}{n-1})^{n-1}} = \frac{x_0}{y_0}$, not what you have – Zach Effman Mar 17 '16 at 16:43
• @ZachEffman I don't think so. Kindly see the steps again. – Hisenberg Mar 17 '16 at 16:44

In general $(ab)^n$ = $a^n b^n$ and $(a^n)^m = a^{nm}$. Therefore raising both sides to the power $n-1$ yields
$$\left(\frac{\cos\alpha}{\sin\alpha}\right)^{n-1} = \left(\left(\frac{b}{a}\right)^{n} \left(\frac{x_0}{y_0}\right)^\frac{1}{n-1}\right)^{n-1} =\left(\frac{b}{a}\right)^{n(n-1)} \left(\frac{x_0}{y_0}\right).$$
• the step where he equated $m_1 = m_2$ – Hisenberg Mar 17 '16 at 16:59