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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?

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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.

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$$\sqrt[-n]{x}=x^{1/-n}$$ 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 $n\in\mathbb Z^-.$

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