<s>[INST] Solve for a in Exponential equation Is it possible to solve for $a$ in the following equation: $a^\alpha=b^\alpha-a$? Currently, I have resorted to using Excel to approximate $a$ (I am given values for $b$ and $\alpha$), but am wondering if it is possible to pinpoint $a$ exactly.   
 [/INST] Although it can not be solved generally, it can be solved if $\alpha \in \{ -1,0, \frac 12,1,2 \}$
Let's consider these in turn:



*

*$\alpha = -1$


$a^{-1} =b^{-1}-a$
$b =a-a^2b$
$a^2b-a+b=0$
$a= \displaystyle \frac {1 \pm \sqrt{(-1)^2-4(b)(b)}}{2b}$
$a= \displaystyle \frac {1 \pm \sqrt{1-4b^2}}{2b}$



*$\alpha = 0$


$a^0 =b^0-a$
$1 =1-a$
$a=0$



*$\alpha = \frac 12$


$a^{\frac 12} =b^{\frac 12}-a$
Let $x=a^{\frac 12} \Rightarrow x^2=a$
$x =b^{\frac 12}-x^2$
$x^2+x-b^{\frac 12}=0$
$x= \displaystyle \frac {-1 \pm \sqrt{1+4b^{\frac 12}}}{2}$
$a= \displaystyle \left(\frac {-1 \pm \sqrt{1+4b^{\frac 12}}}{2}\right)^2$



*$\alpha = 1$


$a^{1} =b^{1}-a$
$a =b-a$
$2a=b$
$a= \displaystyle \frac b2$



*$\alpha = 2$


$a^{2} =b^{2}+a$
$a^2-a-b^2=0$
$a= \displaystyle \frac {1 \pm \sqrt{(-1)^2-4(-b^2)}}{2}$
$a= \displaystyle \frac {1 \pm \sqrt{1+4b^2}}{2}$
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