If $x\in\mathbb R$, solve $x^{\lfloor x\rfloor}=\frac{9}{2}$, where $\lfloor x\rfloor$ is the integer part of $x$.

Question

If $x\in\mathbb R$, solve $x^{\lfloor x\rfloor}=\frac{9}{2}$, where $\lfloor x\rfloor$ is the integer part of $x$. Of course, $x=\lfloor x\rfloor+\{x\}$, where $\{x\}$ is the fractional part of $x$. Hence $0\le\{x\}<1$.

I've added all the meanings of denotions so that everyone can see. It's a consensus to mark all these this way, though.

No calculus can be used in the solution.

The usual ways I solve such problems haven't helped me this time. So some observations would be great. Thanks.

Answer 1

Well, notice that any $x\in[1,2]$ is too small, and any $x\in[3,4]$ is too big. You can replace $x=2+z$, where $z$ is the fractional part. Then $$ (2+z)^2=z^2+4z+4=9/2. $$ Can you conclude?

Answer 2

If $1 \le x < 2$, then $x^{\lfloor x\rfloor} = x < 2$.
If $3 \le x < 4$, then $x^{\lfloor x\rfloor} = x^3 \ge 27$.

Therefore...