Regarding dimensions As the title states, my question is regarding dimensions of a physical quantity.
My question is, can the exponent of dimension of a physical quantity be a fraction?
I came up with this question while solving this problem.
Problem. The potential energy of a particle is $U.$ From a fixed point, if the distance of the particle is $x,$ then $$U=\frac{A\sqrt{x}}{x+B},$$ where $A,B$ are constants. Find $\dim AB.$
Solution. We know that $\dim U=\text M\text L^2\text T^{-2}.$
And $\dim x=\text L.$
This yields -
$$\begin{align*}
U&=\frac{A\sqrt{x}}{x+B}
\\\implies U(x+B)&=A\sqrt{x}
\\\implies \dim U\times\dim x+\dim U\times\dim B&=\dim A\times\dim \sqrt{x}
\\\implies \text M\text L^3\text T^{-2}+\text M\text L^2\text T^{-2}\times \dim B&=\dim A\times L
\\\implies \text M\text L^2\text T^{-2}+\text M\text L^1\text T^{-2}\times \dim B&=\dim A
\\\implies \dim B&=L,\text{ as on LHS, one quantity has dimension }\text M\text L^2\text T^{-2}.
\\\implies \dim A&=\text M\text L^2\text T^{-2}
\\\implies \dim AB&=\text M\text L^3\text T^{-2} \square
\end{align*}$$
But, is it possible, that $\dim\sqrt{x}=L^{\frac{1}{2}}?$
Then the calculation yields
$$\dim A=\text M\text L^{\frac{5}{2}}\text T^{-2},\dim B=L,\dim AB=\text M\text L^{\frac{7}{2}}\text T^{-2}.$$
Which one is correct?
 A: The dimension of $\sqrt{x}$ is $L^{1/2}$, as you say. Therefore your first calculation is not correct (fourth line). 
A: Yes there are various instances where the square root of a fundamental SI unit shows up in the dimensions of a quantity.
The first one that springs to mind is the noise figure of an electronic device. Noise is often a function of frequency, and the noise of an opamp might be given as 5 nV/$\sqrt{Hz}$. This makes sense because noise is related to bandwidth, and it's the power in the signal (which scales with voltage squared) that is the thing that matters. See for example this question and related answers.
As @user40085 pointed out, you made a mistake in the fourth line of your calculation. You can see by inspection that the units of $B$ must be $L$, since $x+B$ is in the denominator. That means $x$ and $B$ must have the same dimensions. Then A has dimensions of $U L/\sqrt{L}$ and your second result follows immediately.
A: Why do you think there is a problem with fractional dimensions as far as I can tell the only two things to remember about dimensions are


*

*Two quantities are added, subtracted, compared or equated only if they have same dimensions 

*Arguments of trigonometric, exponential, logarithmic function should be dimensionless 


And that's it . All question related to dimensions can be done by keeping these two things in mind!
