rational expression
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\(\displaystyle x^{\frac{5}{2}} \ = \ 252\)
The lefthand side (LHS) is not a "rational expression".
It is a "radical expression".
We can say that the exponent itself is a rational expression, but this equation is defined by a radical expression.
It's the same as:
\(\displaystyle \sqrt{x^5} \ = \ 252\)
This equation has only one Real solution because x^5 has only one (principal) square root.
So, the solution "set" will have only one member.
There are some Complex solutions with an imaginary part. Do you know if you're trying to find those, too? (Have you been studying DeMoivre's Theorem?)
To find the Real solution, you could use a property of exponents to change the exponent from 5/2 to 1.
\(\displaystyle (x^n)^m \ = \ x^{n \cdot m}\)
In other words, if n is 5/2, what does m need to be, so that their product is 1?
Raise both sides of the given equation to that value, and the LHS will simplify to x.
Does this make sense?
Stick 252^m into a scientific calculator, to get a decimal approximation for x.
If you're still confused, let us know.
\(\displaystyle x^{\frac{5}{2}} \ = \ 252\)
The lefthand side (LHS) is not a "rational expression".
It is a "radical expression".
We can say that the exponent itself is a rational expression, but this equation is defined by a radical expression.
It's the same as:
\(\displaystyle \sqrt{x^5} \ = \ 252\)
This equation has only one Real solution because x^5 has only one (principal) square root.
So, the solution "set" will have only one member.
There are some Complex solutions with an imaginary part. Do you know if you're trying to find those, too? (Have you been studying DeMoivre's Theorem?)
To find the Real solution, you could use a property of exponents to change the exponent from 5/2 to 1.
\(\displaystyle (x^n)^m \ = \ x^{n \cdot m}\)
In other words, if n is 5/2, what does m need to be, so that their product is 1?
Raise both sides of the given equation to that value, and the LHS will simplify to x.
Does this make sense?
Stick 252^m into a scientific calculator, to get a decimal approximation for x.
If you're still confused, let us know.