Concerning Rules of Exponents & Absolute Value

Question

I understand that one of the accepted definitions of the absolute value function is $\left| x \right| = \sqrt{x^2}$.

However, I do not understand why if I substitute $-5$ in for $x$ that I can't do the following using rules of exponents:

$\left|-5\right| = \sqrt{(-5)^2} = \left[(-5)^2\right]^\frac 12 = (-5)^\frac 22 = (-5)^\frac 11 = -5$.

Clearly we know that the end result isn't right, but I can't find any logical problems with my reasoning. Can someone shed some light onto the situation for me?

Thanks!

Answer 1

The rule you're using: $$(a^b)^c = a^{bc} $$ applies for all $b, c \in \mathbb R$ if and only if $a \geq 0$. If $b, c \in \mathbb Z$, then it also holds for $a \lt 0$.

So $$((-5)^2)^{\large\frac 12} = (25)^{1/2} = 5$$

Answer 2

$(x^{2})^{\frac{1}{2}}=x$ is only valid for $x\ge0$. If $x<0$ notice that:

$(x^{2})^{\frac{1}{2}}=e^{\frac{1}{2}\log(x^{2})}=e^{\frac{1}{2}\log(\lvert x\rvert)^{2}}=e^{\log(\lvert x\rvert)}=\lvert x\rvert\neq x$

Answer 3

A simple idea may be : $$\sqrt{(-5)^2} = \left[(-5)^2\right]^\frac 12 = (-5)^\frac 22 = ((-5)^\frac 12)^2 = (\sqrt{-5})^2$$ and there is no meaning for $\sqrt{-5}$ when we talk about real numbers.