# Solving for $a$ (log probably required)

I need to get $c$ in terms of $a$.

Both $c, a\in \Bbb R^+$ $$c^{2a} = c/a$$

I'm pretty sure that log is required to solve this, but I'm not quite sure how to approach it.

• How about dividing by $c$ then taking the appropriate power? – Jonathan Y. Aug 29 '13 at 20:43
• @JonathanY. Oh, that seems like it might make more sense. It just seemed to me like log would be needed, because of the powers – Alex Aug 29 '13 at 20:46

Since $c>0$, we may divide both sides by $c$ to obtain: $$c^{2a-1} = \frac1a$$ To undo the exponent, we raise both sides of the equation to the power of $\dfrac{1}{2a-1}$ to obtain: $$c= \left(\frac1a\right)^{\dfrac{1}{2a-1}}$$
• It should probably be pointed out that this doesn't work if $a=\frac{1}{2}$; however, in that case the equation is $c=c/a$, so that either $a=1$ or $c=0$. – Nick Peterson Aug 29 '13 at 20:46
• @NicholasR.Peterson If $a=\frac{1}{2}$, how can $a=1$? – Ryan Aug 29 '13 at 22:07
• Well, it can't. So that tells you that $c=0$. :-P – Nick Peterson Aug 29 '13 at 22:08