if $f(g(x))=x^3$ and $g(f(x))=x^4$ find $f(x)$ and $g(x)$ I am taking algebra two and questions like this popped into my head so I would like to learn how to solve them, I am currently working on composing functions.
 A: We have $$ f(x^4) = f(g(f(x)) = (f(x))^3 $$
Let $F(y) := \log f(e^y)$. We have
$$ F(4y) = \log f(e^{4y}) = \log \big((f(e^y))^3\big) = 3\log f(e^y) = 3F(y) $$
You can easily construct such functions. You can take any function $F : (-4,-1]\cup[1,4) \rightarrow \mathbb R$ and then define $F(4^k y) = 3^k F(y)$ for $k\in\mathbb Z$, which will define $F: (-\infty,0)\cup(0,\infty)\rightarrow \mathbb R$, and finally put $F(0)=0$.
You can take then $$ f(x) = \exp F(\log x)$$
which will define $f: (0,\infty) \rightarrow \mathbb (0,\infty)$.
If this function turns out to be invertible (which happens if $F:\mathbb R\rightarrow\mathbb R$ is invertible), then you can define $g(x) = f^{-1}(x^3)$
Example:
$$ F(y) = \left\{\begin{array}{ll} y^{\log_4 3} & \text{for }y>0 \\ 0 & \text{for }y=0 \\ -(-y)^{\log_4 3} &\text{for }y<0 \end{array}\right.$$ $$ f(x) = \left\{\begin{array}{ll} \exp\big((\log x)^{\log_4 3}\big) & \text{for }x>1 \\ 1 & \text{for }x=1 \\ \exp\big(-(-\log x)^{\log_4 3}\big) &\text{for }0<x<1 \end{array}\right.$$ $$ g(x) = \left\{\begin{array}{ll} \exp\big((3\log x)^{\log_3 4}\big) & \text{for }x>1 \\ 1 & \text{for }x=1 \\ \exp\big(-(-3\log x)^{\log_3 4}\big) &\text{for }0<x<1 \end{array}\right.$$
