To find, wether '1' lies in the range of f, where $f(x)=[ln(\frac{7x-x^2}{12})]^\frac{3}{2}$?

Question

$f(x)=[ln(\frac{7x-x^2}{12})]^\frac{3}{2}$,

For the given function, the question is whether, f(x) can equal 1 for some real value of x?

Answer 1

The tactic is to make things simpler.

If there is $x$ such that $f(x)=[ln(\frac{7x-x^2}{12})]^\frac{3}{2}=1$ then we have also

$ln(\frac{7x-x^2}{12})=1$, that is, $\dfrac{7x-x^2}{12}=e$ or even simpler

$x^2-7x+12e=0$

You just have to check wether this equation has real solutions.