Prove that $f(x)=e^x\ln⁡ x$ attains every real number as its value exactly once how can I prove that the function
$f(x)=e^x\ln x$

attains every real number as its value exactly once. (by proving that it's a  monotonic continuous function)?
The function is defined for $x&gt;0$; it should be easy to prove that
$$
\lim_{x\to0^+}f(x)=-\infty,\qquad
\lim_{x\to\infty}f(x)=\infty
$$
so the range of the function is $\mathbb{R}$ because…

The derivative is
$$
f'(x)=\frac{e^x}{x}(1+x\ln x)
$$

Consider $g(x)=1+x\ln x$; then
$$
g'(x)=1+\ln x
$$
that vanishes for $x=e^{-1}$ and $g(e^{-1})=1-e^{-1}&gt;0$.

Can you conclude?
Take it's derivative: $f'(x)=e^x \ln x + \frac{e^x}{x}$ and note that it is always strictly greater than 0.

This shows that it is strictly increasing thus injective, so if every real number is reached, than also exactly once.

Now because $\lim_{x\rightarrow 0} f(x) = -\infty$ and $\lim_{x\rightarrow \infty} f(x) = \infty$, every real number is reached because of the mean value theorem.