Evaluate $(b^2)^{\frac{1}{2}}$ for $b=-1$.
If you substitute $b=-1$ first and then simplify you get $((-1)^2)^{\frac{1}{2}}=(1)^{\frac{1}{2}}=1$ but if you simplify first and then substitute you get $b^1=(-1)^1=-1$.
What am I missing?
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1The problem is that the exponential identity $(a^b)^c=a^{bc}$ fails if $a$ is negative, which it is here. – Parcly Taxel Oct 18 '16 at 04:24
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In general, the definitions an laws of eponents don't apply for negative bases for precisely these reasons. Also note $(-1)^{1/2} $ isn't defined. Nor is any $(-1)^x $ for irrational x (the most common definition for $b^x$ is $e^{x\ln b}$ which isn't possible for negative b). So basically, if b is negative $(b^x)^y=(b^y)^x=b^{xy} $ only applies if x, y are rational with odd denominators. Which is too restrictive to be practical. However $((-1)^2)^{1/2}=(1)^{1/2}=1$ is well defined and acceptable $((-1)^2)^{1/2}=(-1)^1$ is not. – fleablood Oct 18 '16 at 05:26