# Negative solution to an equation with variable on the index of a root

I was working on the following problem on exponential equations:

Solve for $$x$$: $$\sqrt{8^{x-1}}\cdot\sqrt[x+1]{4^{2x-3}}=\sqrt[6]{2^{5x+3}}$$

(The second root has an index of $$x+1$$) And I got $$S=\{2,-6\}$$ for a solution, but the book's solution says that $$-6$$ isn't a valid solution, because then $$x+1$$ would be negative, and the index of a root must be a natural number. But $$\sqrt[-n]{x}=x^{1/-n}=x^{-1/n}=\left(\frac{1}{x}\right)^{1/n}=\sqrt[n]{\frac{1}{x}}$$

I understand that we conventionally don't put negative numbers on the index of a root because there's no good reason to do it, but does that really mean that $$-6$$ is not a solution to this equation? Shouldn't $$\sqrt[n]{a^b}=a^{b/n}$$ in all cases?

This is just a matter of convention, and different people can have different conventions. Usually, $$\sqrt[n]{x}$$ is taken to be synonymous with $$x^{1/n}$$ and so it is defined whenever $$x^{1/n}$$ would be. Apparently your book uses a different convention where $$\sqrt[n]{x}$$ is defined only when $$n$$ is a positive integer.
$$\sqrt[-n]{x}=x^{1/-n}$$ I understand that we conventionally don't put negative numbers on the index of a root because there's no good reason to do it,
In real analysis for nonnegative $$x,$$ $$\sqrt[m]{x}:=x^{1/m}$$ conventionally means the nonnegative (i.e., principal) $$n$$th root of $$x.$$ On the other hand, $$\sqrt[-3]{x}$$ is not meaningful, unless the author has specially defined what $$n$$th root means for $$n\in\mathbb Z^-.$$