If the domain of $f(x)$ is $(-3, 1)$, then what is the domain of $f(\ln x)$? [closed]

Question

I need a clear explanation for this question:

If the domain of $f(x)$ is $(-3, 1)$ then the domain of $f(\ln x)$ is ...

a) $\;(e^{-1}, e^3)$
b) $\;(0, \infty)$
c) $\;(1, \infty)$
d) $\;(e^{-3}, e^1)$

As we know, any logarithmic function has a domain of $(0, \infty)$, but the answer suggested that the domain should be $(e^{-3}, e^1)$. Why is this the case?

Answer 1

I will just clear your doubt and you can solve it on your own.
The domain of $ln(x)$ is not equal to the domain of the function $f(ln(x))$

Now you can try it.

Hint : $lnx$ should have the values -3,1

Solution:

Given $f(x)$ has a domain (-3,1), $f(x)$ is valid only when the values of x are in the range (-3,1).
So for the function $f(ln(x))$ to be valid $lnx$ should have the values in the range (-3,1) therefore the domain of the function $f(ln(x))$ should be $(e^{-3},e^{1})$

Answer 2

Since the domain of $f(x)$ is $-3$ to $1$, the function $\ln(x)$ must give an output from $-3$ to $1$.

To achieve this condition, $\ln(x)$ must have the domain $(e^{-3},e^1)$. This is a simple solution to your problem.

The domain of $\ln$ is $(0,\infty)$ but it is not the same domain of $f(\ln(x))$

Please try asking more higher standard questions in this site, as this question goes against the websites policy. This is just being told in a friendly manner.