What is the solution for $x^{\frac {10x}{3} }=\frac {243} {32}$? I understand that $243$ is $3^5$ and $32$ is $2^5$, but I don't know where to go from there after converting it to logarithm.

As you said, $\frac{243}{32}=(\frac{3}{2})^5$. So why not try $x=\frac{3}{2}$? Indeed, we have $\frac{10\cdot \frac{3}{2}}{3}=5$ which makes $x=\frac{3}{2}$ a solution. We will now Show that there are no other solutions.
Obviously, we cannot have $0<x<1$ because that would imply $x^{\frac{10x}{3}}<1$. Furthermore, if we had $x<-1$, it would follow that $x^{\frac{10x}{3}}<1$ as well (if the term even was defined). And then, $x$ can't be between $0$ and $-1$ because that would make the exponent no whole number - but powers with this kind of exponent are only defined for positive reals as the base. So we have $x>1$. Thus, we consider the function $x^{\frac{10x}{3}}$ on the interval $(1, \infty)$. The derivative of this function is $$(x^{\frac{10x}{3}})'=(e^{ln(x)\cdot \frac{10x}{3}})'=e^{ln(x)\cdot \frac{10x}{3}}\cdot (ln(x)\cdot \frac{10}{3})\cdot (\frac{1}{x}\cdot \frac{10x}{3})=\frac{100}{9}\cdot e^{ln(x)\cdot \frac{10x}{3}}\cdot ln(x)>0$$
for all $x\in (1, \infty)$. So our function is strictly monotonically increasing on the interval $(1, \infty)$ and thus there can only be this one solution we discovered above.
