Why does an equation in the form Z = ax^n, where n is a non-integer value greater than 1 (like 1.7) not have?

Question

a solution?

Stupid question, but I can't really think of why.

When solving for x

Answer 1

I am assuming Z and a are given and you are solving for x. Then there may or may not be a solution. Here is one which you can solve:

2 = x^(1.5),

which implies x = 2^(2/3). The problem arises especially when you write a negative number for Z and a positive number for a, in which case you have
x^n = negative.
Now if n is an odd integer then you can find x such that x^n is negative but if n is an even number or if n is a rational which can not be written as a quotient of two odd integers, then that equation can not have a solution as the negative numbers only have odd (3rd, 5th,...) roots but not even roots (square root, 4th root, ...).

Answer 2

I am not sure what you mean by "a solution". Suppose n is 1.5 or 3/2 and x = 4. then x^n = 4^(3/2) = (64)^(1/2) = 8

so Z = 8a

Is that what you mean?

Answer 3

Why...I know why.

Because the NY Mets are in first place and the football season is just around the corner!

Answer 4

Rearrange to

Z/a = X^n

Now if Z/a is defined and non-zero then

ln(Z/a) = n ln(X)

ln(Z/a)/n = ln(X)

X = exp(ln(Z/a)/n)

There is a solution as long as both Z and a are non-zero