If $ f(x) = \ln(x) $, $ g(x)=x^{2/3} $ and $ h(x)=e^x $, then what is $ f(g(h(x)))$? If $ f(x) = \ln(x) $, $ g(x)=x^{2/3} $ and $ h(x)=e^x $, then what is $ f(g(h(x)))$? What is domain of this function?
 A: $$g(h(x))=\left (h(x) \right )^{2/3}=\left (e^x \right )^{2/3}$$
$$f(g(h(x)))=\ln (g(h(x))= \ln \left (e^x \right )^{2/3}=\frac {2x}{3}$$
A: The domain of both $g(x) = x^{2/3}$ and $h(x) = e^x$ is the set of all real numbers $\mathbb{R}$. What you need to be careful about is the function $f(x) = \ln(x)$, whose domain is the positive real numbers $\mathbb{R}_{>0}$.
The input that we feed to $f(x)$ thus needs to be positive, and since we are dealing with $f(g(h(x)))$, the input $g(h(x))$ needs to be positive for the expression to make sense.
When is $g(h(x))$ positive? Well, $h(x)$ is the exponential function and is always positive. You feed this to $g(x) = x^{2/3}$, which is also always positive. In conclusion, no matter what real number $x$ you feed into $g(h(x))$, the output will be positive, and $g(h(x))$ will make sense as an argument to $f(x)$.
Thus, the domain of $f(g(h(x)))$ is $\mathbb{R}$, the set of all real numbers.
A: $$f(g(h(x)))=f(g(e^x))=f(e^{\frac{2x}{3}})=\frac{2x}{3}$$
with domain $D_f=\mathbb{R}$.
A: $f(g(h(x)))$ is a composite function,so start from the inside :  $h(x)=e^x$
$g(h(x))=g(e^x)=({e^x})^\frac{2}{3}$
So, $f(g(h(x)))=f(({e^x})^\frac{2}{3})=ln(({e^x})^\frac{2}{3})$
Therefore, $Dom(f(g(h(x)))):({e^x})^\frac{2}{3}>0\implies x\in\mathbb{R}$
( since $e^x$ is already greater than zero for all x $\in\mathbb{R}$ )
OR you can simplify $ln(({e^x})^\frac{2}{3})= {\frac{2}{3}}x$
so the domain also is $\mathbb{R}$
