What is the right way to calculate a power? I noticed that there are two solutions for $(-1)^{14/2}$:


*

*$((-1)^{14})^{1/2} = 1$

*$(-1)^{14/2}=(-1)^7=-1$


What am I doing wrong?
 A: The root of the problem is that $\sqrt{ 1\;}=+1$, unambiguously by definition, but $1^\frac12=(e^{2\pi i k})^{\frac12}=e^{\pi i k}\;\;\forall k\in\mathbb Z$ is not, it results to $\pm1$ depending on $k$ being odd or even.
Taking roots, you have to choose a branch, like the commonly accepted branch that $\sqrt x \ge 0$ for $x\ge0$. We could just as well have chosen the other branch, but you get in big trouble if you mix those.
A: (One of) the rigourous definitions, and the simplest, of $\;a^{\tfrac{14}2}$ is it's $\;\mathrm e^{\tfrac{14}2\ln a}$ so that it supposes $a>0$ – and avoids this kind of misbehaviour. $a^x$ for a negative $a$ is defined only for integer exponents, because it has an intuitive meaning.
A: The domain of the function $f(x) = (-1)^x$ is the set of all integers. If $x$ is not an integer, the function returns a non-real value.
Specifically, we cannot use the standard rules of exponentation for this problem because the base is negative. As @hardmath has clarified, we should "force" an integer exponent on a value, as fractional exponents are not allowed. Thus, the second solution given in the original post is correct, and the first is invalid.
A: You're not doing anything wrong, per se; the issue is that in order to define square root as a single-valued function on $\mathbb{R}_{\geq 0}$, we (somewhat arbitrarily) choose the positive value.  So both $-1$ and $1$ square to $1$, but the square root of $1$ is only $1$, not $-1$.  This becomes a bit more transparent if we replace $14$ by $2$:
$$
[(-1)^2]^{1/2} = 1^{1/2} = 1 \not= -1 = (-1)^1 = (-1)^{2/2}
$$
Accordingly, the identity $(a^b)^c = a^{bc}$ does not hold generally for negative $a$, if $b$ and $c$ are not both integers.
A: In real numbers, the standard definition of rational exponents only permits fully reduced fractions in the exponent. Example definition from Sullivan's College Algebra:

Definition. If $a$ is a real number and $m$ and $n$ are integers containing no common factors, with $n \geq 2$, then $$a^{m/n}=\sqrt[n]{a^m}=\left(\sqrt[n]a\right)^m$$ provided that $\sqrt[n]{a}$ exists.
We have two comments about this equation:

*

*The exponent $\frac mn$ must be in lowest terms and $n$ must be positive.

*In simplifying the rational expression $a^{m/n}$, either $\sqrt[n]{a^m}$ or $\left(\sqrt[n]a\right)^m$ may be used, the choice depending on which is easier to simplify. Generally, taking the root first, as in $\left(\sqrt[n]a\right)^m$, is easier.


The first suggested transformation, $((-1)^{14})^{1/2} = 1$, is therefore invalid under this definition, because it disallows separating the numerator and denominator when they have a common factor.
Therefore only the second transformation, $(-1)^{14/2}=(-1)^7=-1$, which reduces the rational exponent first, is valid under a real-number definition like this.
A: Formally speaking, $a^{1/2}$ is defined to be such a special number $b$ that $b^2 =a$. That means $1^{1/2}$ can formally be either $+1$ or $-1$, we just choose the $+1$ for convenience...
A: The standard rules for exponentiation in $\mathbb R$ only always work for positive real number bases, and
$((-1)^{14})^{1/2} = 1 \ne (-1)^{\frac{14}{2}}$
is an example of that.
The rules of order of operation reduce 
$((-1)^{14})^{1/2}$
to 1.
To claim that this is not true would mean that the rules of order of operation are faulty. Because of that, I disagree with any one who wants the answer to be $-1$.
The Standard Rules
If $x,y,m,n \in \mathbb R$ with $x,y \gt 0$, then


*
    
* $x^m x^n = x^{m+n}$
    
* $(x^m)^n = x^{mn}$
    
* $(xy)^n = x^n y^n$
    
* $x^0 = 1$
    
* $x^1 = x$
    
* $\sqrt[n]{x^m} = x^\frac{m}{n}$



When $x\lt 0$, then the standard rules do not always work. It's possible to come up with special cases that do work; but, in my mind, it's not worth the effort.
