text: # Conditions of exponential functions
  
  Okay, is the above statement correct? Because if I put $$x=\frac{1}{2}$$ then $$f(x)$$ will have two values. So will that remain a function anymore?
  
  • Working with exponential functions (as defined above), we only consider the principal root. For example, $a^{\frac{11}{4}}$ could be interpreted as $\sqrt[4]{a^{11}}$. This is perfect for defining a function. But you can't use this notion everywhere. Consider $(-3)^2=9$, which means $(-3)^{2\cdot\frac{1}{2}}=9^{\frac{1}{2}}$, so $(-3)^1=-3=3$? This is wrong of course. – Corellian Oct 19 '18 at 5:39
  
  Indeed, the unrestricted inverse of any non-injective function is multivalued, but this is fixed by simply restricting the co-domain to some interval. Take $$f(x)=y=x^2$$ for example. It's inverse satisfies $$f^{-1}(y) = x$$. Therefore, if we just solve our original function for $$x$$, we should have our answer.
  $$x^2 = y \\ x^2 - y = 0 \\ (x+y^{\frac{1}{2}})(x-y^{\frac{1}{2}})=0 \\ f^{-1}(y) = x = \pm y^{\frac{1}{2}}$$
  Clearly, this is multivalued, and cannot be a function, so by convention, we define the principal square root function to be positive solution to be the inverse of $$f(x) = y = x^2$$.
  In conclusion, the statement is correct because $$f(x)=a^x$$ only has one value at $$f(\frac{1}{2})$$.