When is $(a^b)^c $ = $a^{bc}$ true?

Question

I know that in some cases this rule is not true. For example

$$((-1)^2)^\frac{1}{2}\ne(-1)^{(2\cdot\frac{1}{2})}$$

So when is this rule true ?

Answer 1

It is true when $a>0$ and $b, c \in \mathbb R$. In this situation we have a canonical choice for a value for $a^b$, given by $e^{b \ln a}$, using the principal branch of the natural logarithm on positive real numbers (that is, the one that gives $\ln a \in \mathbb R$).

For other values of $a$ there is no natural choice for $\ln a$, so $a^b$ has multiple equally valid answers. What is true is that each value of $a^{bc}$ is equal to a value for $(a^b)^c$, but not conversely. For instance, in the example you gave, $a=-1, b = 2, c = \frac{1}{2}$, the values of $a^{bc}$ are

$$(-1)^1 = e^{\ln(-1)} = e^{\pi i + 2\pi i n} = -1,$$

and the values of $(a^b)^c$ are

$$( (-1)^2)^{\frac{1}{2}} = 1^{\frac{1}{2}} = e^{\frac{1}{2} \ln(1)} = e^{\frac{1}{2} 2\pi i n} = \pm 1.$$

Answer 2

If $b,c$ are integer numbers than the rule $(a^b)^c=a^{bc}$ is true for all $a \in \mathbb{R}$. If at least one of the two exponent is a real non integer number than the rule is true only for non negative basis $a$.

Answer 3

The main two cases where this rule holds are

1. When $a$ is a positive real (and $b, c$ are real).
2. When $b$ and $c$ are both integers.

In the latter case it holds not only for multiplication of numbers, but in any group.

The prototypical case of exponentiation, where base and exponent are both positive integers, is a special case of each of these situations.